Program to check for a Valid IMEI Number Last Updated : 28 Jul, 2022 Comments Improve Suggest changes Like Article Like Report International Mobile Equipment Identity (IMEI) is a number, usually unique, to identify mobile phones, as well as some satellite phones. It is usually found printed inside the battery compartment of the phone, but can also be displayed on-screen on most phones by entering *#06# on the dialpad, or alongside other system information in the settings menu on smartphone operating systems. The IMEI number is used by a GSM network to identify valid devices and therefore can be used for stopping a stolen phone from accessing that network. The IMEI (15 decimal digits: 14 digits plus a check digit) includes information on the origin, model, and serial number of the device. The IMEI is validated in following steps: Starting from the rightmost digit, double the value of every second digit (e.g., 7 becomes 14).If doubling of a number results in a two digits number i.e greater than 9(e.g., 7 × 2 = 14), then add the digits of the product (e.g., 14: 1 + 4 = 5), to get a single digit number.Now take the sum of all the digits.Check if the sum is divisible by 10 i.e.(total modulo 10 is equal to 0) then the IMEI number is valid; else it is not valid. Example: Input IMEI : 490154203237518 Output : Since, 60 is divisible by 10, hence the given IMEI number is Valid. Implementation: C++ // C++ program to check whether the // given EMEI number is valid or not. #include<bits/stdc++.h> using namespace std; // Function for finding and returning // sum of digits of a number int sumDig(int n) { int a = 0; while (n > 0) { a = a + n % 10; n = n / 10; } return a; } bool isValidIMEI(long n) { // Converting the number into // String for finding length string s = to_string(n); int len = s.length(); if (len != 15) return false; int sum = 0; for(int i = len; i >= 1; i--) { int d = (int)(n % 10); // Doubling every alternate digit if (i % 2 == 0) d = 2 * d; // Finding sum of the digits sum += sumDig(d); n = n / 10; } return (sum % 10 == 0); } // Driver code int main() { // 15 digits cannot be stored // in 'int' data type long n = 490154203237518L; if (isValidIMEI(n)) cout << "Valid IMEI Code"; else cout << "Invalid IMEI Code"; return 0; } // This code is contributed by Yash_R Java // Java program to check whether the // given EMEI number is valid or not. import java.io.*; class IMEI { // Function for finding and returning // sum of digits of a number static int sumDig(int n) { int a = 0; while (n > 0) { a = a + n % 10; n = n / 10; } return a; } static boolean isValidIMEI(long n) { // Converting the number into String // for finding length String s = Long.toString(n); int len = s.length(); if (len != 15) return false; int sum = 0; for (int i = len; i >= 1; i--) { int d = (int)(n % 10); // Doubling every alternate digit if (i % 2 == 0) d = 2 * d; // Finding sum of the digits sum += sumDig(d); n = n / 10; } return (sum % 10 == 0); } // Driver code public static void main(String args[]) throws IOException { // 15 digits cannot be stored in 'int' data type long n = 490154203237518L; if (isValidIMEI(n)) System.out.println("Valid IMEI Code"); else System.out.println("Invalid IMEI Code"); } } Python3 # Python3 code to check whether the # given EMEI number is valid or not # Function for finding and returning # sum of digits of a number def sumDig( n ): a = 0 while n > 0: a = a + n % 10 n = int(n / 10) return a # Returns True if n is valid EMEI def isValidEMEI(n): # Converting the number into # String for finding length s = str(n) l = len(s) # If length is not 15 then IMEI is Invalid if l != 15: return False d = 0 sum = 0 for i in range(15, 0, -1): d = (int)(n % 10) if i % 2 == 0: # Doubling every alternate digit d = 2 * d # Finding sum of the digits sum = sum + sumDig(d) n = n / 10 return (sum % 10 == 0) # Driver code n = 490154203237518 if isValidEMEI(n): print("Valid IMEI Code") else: print("Invalid IMEI Code") # This code is contributed by "Sharad_Bhardwaj". C# // C# program to check whether the // given EMEI number is valid or not. using System; class GFG { // Function for finding and // returning sum of digits // of a number static int sumDig(int n) { int a = 0; while (n > 0) { a = a + n % 10; n = n / 10; } return a; } static Boolean isValidIMEI(long n) { // Converting the number into // String for finding length String s = n.ToString(); int len = s.Length; if (len != 15) return false; int sum = 0; for (int i = len; i >= 1; i--) { int d = (int)(n % 10); // Doubling every alternate // digit if (i % 2 == 0) d = 2 * d; // Finding sum of the digits sum += sumDig(d); n = n / 10; } return (sum % 10 == 0); } // Driver code public static void Main() { // 15 digits cannot be stored in // 'int' data type long n = 490154203237518L; if (isValidIMEI(n)) Console.Write("Valid IMEI Code"); else Console.Write("Invalid IMEI Code"); } } // This code is contributed by parashar. JavaScript <script> // javascript program to check whether the // given EMEI number is valid or not. // Function for finding and returning // sum of digits of a number function sumDig(n) { let a = 0; while (n > 0) { a = a + n % 10; n = parseInt(n / 10, 10); } return a; } function isValidIMEI(n) { // Converting the number into // String for finding length let s = n.toString(); let len = s.length; if (len != 15) return false; let sum = 0; for(let i = len; i >= 1; i--) { let d = (n % 10); // Doubling every alternate digit if (i % 2 == 0) d = 2 * d; // Finding sum of the digits sum += sumDig(d); n = parseInt(n / 10, 10); } return (sum % 10 == 0); } // 15 digits cannot be stored // in 'int' data type let n = 490154203237518; if (isValidIMEI(n)) document.write("Valid IMEI Code"); else document.write("Invalid IMEI Code"); // This code is contributed by vaibhavrabadiya117. </script> OutputValid IMEI Code Time complexity : O(n log10 n)Auxiliary Space: O(n) Comment More infoAdvertise with us Next Article Analysis of Algorithms A Anshika Goyal Improve Article Tags : Misc Strings Technical Scripter DSA Practice Tags : MiscStrings Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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