Check if a number is multiple of 9 using bitwise operators

Given a number n, write a function that returns true if n is divisible by 9, else false. The most simple way to check for n's divisibility by 9 is to do n%9. 
Another method is to sum the digits of n. If sum of digits is multiple of 9, then n is multiple of 9. 
The above methods are not bitwise operators based methods and require use of % and /. 
The bitwise operators are generally faster than modulo and division operators. Following is a bitwise operator based method to check divisibility by 9. 
 

// C++ program to check if a number
// is multiple of 9 using bitwise operators
#include <bits/stdc++.h>
using namespace std;

// Bitwise operator based function to check divisibility by 9
bool isDivBy9(int n)
{
    // Base cases
    if (n == 0 || n == 9)
        return true;
    if (n < 9)
        return false;

    // If n is greater than 9, then recur for [floor(n/9) - n%8]
    return isDivBy9((int)(n >> 3) - (int)(n & 7));
}

// Driver program to test above function
int main()
{
    // Let us print all multiples of 9 from 0 to 100
    // using above method
    for (int i = 0; i < 100; i++)
        if (isDivBy9(i))
            cout << i << " ";
    return 0;
}

// Java program to check if a number
// is multiple of 9 using bitwise operators
import java.lang.*;

class GFG {

    // Bitwise operator based function
    // to check divisibility by 9
    static boolean isDivBy9(int n)
    {

        // Base cases
        if (n == 0 || n == 9)
            return true;
        if (n < 9)
            return false;

        // If n is greater than 9, then
        // recur for [floor(n/9) - n%8]
        return isDivBy9((int)(n >> 3) - (int)(n & 7));
    }

    // Driver code
    public static void main(String arg[])
    {

        // Let us print all multiples of 9 from
        // 0 to 100 using above method
        for (int i = 0; i < 100; i++)
            if (isDivBy9(i))
                System.out.print(i + " ");
    }
}

// This code is contributed by Anant Agarwal.

# Bitwise operator based
# function to check divisibility by 9

def isDivBy9(n):

    # Base cases
    if (n == 0 or n == 9):
        return True
    if (n < 9):
        return False
 
    # If n is greater than 9,
    # then recur for [floor(n / 9) - n % 8]
    return isDivBy9((int)(n>>3) - (int)(n&7))

# Driver code

# Let us print all multiples
# of 9 from 0 to 100
# using above method
for i in range(100):
    if (isDivBy9(i)):
        print(i, " ", end ="")

# This code is contributed
# by Anant Agarwal.

// C# program to check if a number
// is multiple of 9 using bitwise operators
using System;

class GFG {

    // Bitwise operator based function
    // to check divisibility by 9
    static bool isDivBy9(int n)
    {
        // Base cases
        if (n == 0 || n == 9)
            return true;
        if (n < 9)
            return false;

        // If n is greater than 9, then
        // recur for [floor(n/9) - n%8]
        return isDivBy9((int)(n >> 3) - (int)(n & 7));
    }

    // Driver code
    public static void Main()
    {
        // Let us print all multiples of 9 from
        // 0 to 100 using above method
        for (int i = 0; i < 100; i++)
            if (isDivBy9(i))
                Console.Write(i + " ");
    }
}

// This code is contributed by nitin mittal.

<?php
// PHP program to check if a number
// is multiple of 9 using bitwise
// operators

// Bitwise operator based function
// to check divisibility by 9
function isDivBy9($n)
{
    
    // Base cases
    if ($n == 0 || $n == 9)
        return true;
    if ($n < 9)
        return false;

    // If n is greater than 9, 
    // then recur for [floor(n/9) - 
    // n%8]
    return isDivBy9(($n >> 3) - 
                    ($n & 7));
}

    // Driver Code
    // Let us print all multiples
    // of 9 from 0 to 100
    // using above method
    for ($i = 0; $i < 100; $i++)
        if (isDivBy9($i))
            echo $i ," ";
            
// This code is contributed by nitin mittal
?>

<script>
// javascript program to check if a number
// is multiple of 9 using bitwise operators

// Bitwise operator based function
// to check divisibility by 9
function isDivBy9(n)
{

    // Base cases
    if (n == 0 || n == 9)
        return true;
    if (n < 9)
        return false;

    // If n is greater than 9, then
    // recur for [floor(n/9) - n%8]
    return isDivBy9(parseInt(n >> 3) - parseInt(n & 7));
}

// Driver code

    // Let us print all multiples of 9 from
// 0 to 100 using above method
for (i = 0; i < 100; i++)
    if (isDivBy9(i))
        document.write(i + " ");

// This code is contributed by Princi Singh 
</script>

Output: 

0 9 18 27 36 45 54 63 72 81 90 99

Time Complexity: O(log n)

Auxiliary Space: O(logn)
How does this work? 
n/9 can be written in terms of n/8 using the following simple formula. 

n/9 = n/8 - n/72

Since we need to use bitwise operators, we get the value of floor(n/8) using n>>3 and get value of n%8 using n&7. We need to write above expression in terms of floor(n/8) and n%8. 
n/8 is equal to "floor(n/8) + (n%8)/8". Let us write the above expression in terms of floor(n/8) and n%8

n/9 = floor(n/8) + (n%8)/8 - [floor(n/8) + (n%8)/8]/9
n/9 = floor(n/8) - [floor(n/8) - 9(n%8)/8 + (n%8)/8]/9
n/9 = floor(n/8) - [floor(n/8) - n%8]/9

From above equation, n is a multiple of 9 only if the expression floor(n/8) - [floor(n/8) - n%8]/9 is an integer. This expression can only be an integer if the sub-expression [floor(n/8) - n%8]/9 is an integer. The subexpression can only be an integer if [floor(n/8) - n%8] is a multiple of 9. So the problem reduces to a smaller value which can be written in terms of bitwise operators.
 

kartik

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Article Tags :

• Bit Magic
• DSA
• divisibility

Practice Tags :

• Bit Magic