Check if a number is divisible by 23 or not - GeeksforGeeks

17 May, 2021
Given a number, the task is to quickly check if the number is divisible by 23 or not.Examples:  
Input : x  = 46
Output : Yes

Input : 47
Output : No
 
A solution to the problem is to extract the last digit and add 7 times of last digit to remaining number and repeat this process until a two digit number is obtained. If the obtained two digit number is divisible by 23, then the given number is divisible by 23.Approach: 
Extract the last digit of the number/truncated number every time
Add 7*(last digit of the previous number) to the truncated number
Repeat the above three steps as long as necessary.
Illustration:  
 
17043-->1704+7*3 
= 1725-->172+7*5
= 207 which is 9*23, 
so 17043 is also divisible by 23.
 
Mathematical Proof : Let be any number such that =100a+10b+c . Now assume that is divisible by 23. Then 0 (mod 23) 100a+10b+c0 (mod 23) 10(10a+b)+c0 (mod 23) 10+c0 (mod 23)Now that we have separated the last digit from the number, we have to find a way to use it. Make the coefficient of 1. In other words, we have to find an integer such that n such that 10n1 mod 23. It can be observed that the smallest n which satisfies this property is 7 as 701 mod 23. Now we can multiply the original equation 10+c0 (mod 23) by 7 and simplify it: 70+7c0 (mod 23) +7c0 (mod 23) We have found out that if 0 (mod 23) then, +7c0 (mod 23). In other words, to check if a 3-digit number is divisible by 23, we can just remove the last digit, multiply it by 7, and then subtract it from the rest of the two digits.  
 
C++
Java
Python 3
C#
PHP
Javascript
// CPP program to validate above logic#include <iostream>using namespace std; // Function to check if the number is// divisible by 23 or notbool isDivisible(long long int n){     // While there are at least 3 digits    while (n / 100)    {        int d = n % 10; // Extracting the last digit        n /= 10; // Truncating the number         // Adding seven times the last        // digit to the remaining number        n += d * 7;    }     return (n % 23 == 0);} int main(){    long long int n = 1191216;    if (isDivisible(n))        cout << "Yes" << endl;    else        cout << "No" << endl;    return 0;}
// Java program to validate above logicclass GFG{     // Function to check if the// number is divisible by// 23 or notstatic boolean isDivisible(long n){     // While there are at    // least 3 digits    while (n / 100 != 0)    {        // Extracting the last digit        long d = n % 10;                 n /= 10; // Truncating the number         // Adding seven times the last        // digit to the remaining number        n += d * 7;    }     return (n % 23 == 0);} // Driver Codepublic static void main(String[] args){    long n = 1191216;    if(isDivisible(n))        System.out.println("Yes");    else        System.out.println("No");}} // This code is contributed by mits
# Python 3 program to validate above logic # Function to check if the number is# divisible by 23 or notdef isDivisible(n) :     # While there are at least 3 digits    while n // 100 :         # Extracting the last        d = n % 10         # Truncating the number        n //= 10         # Adding seven times the last         # digit to the remaining number        n += d * 7     return (n % 23 == 0) # Driver Codeif __name__ == "__main__" :     n = 1191216     # function calling    if (isDivisible(n)) :        print("Yes")     else :        print("No") # This code is contributed by ANKITRAI1
// C# program to validate// above logicclass GFG{     // Function to check if the// number is divisible by// 23 or notstatic bool isDivisible(long n){     // While there are at    // least 3 digits    while (n / 100 != 0)    {        // Extracting the last digit        long d = n % 10;                 n /= 10; // Truncating the number         // Adding seven times the last        // digit to the remaining number        n += d * 7;    }     return (n % 23 == 0);} // Driver Codepublic static void Main(){    long n = 1191216;    if(isDivisible(n))        System.Console.WriteLine("Yes");    else        System.Console.WriteLine("No");}} // This code is contributed by mits
<?php// PHP program to validate above logic // Function to check if the number// is divisible by 23 or notfunction isDivisible($n){     // While there are at    // least 3 digits    while (intval($n / 100))    {        $n = intval($n);        $d = $n % 10; // Extracting the last digit        $n /= 10; // Truncating the number         // Adding seven times the last        // digit to the remaining number        $n += $d * 7;    }     return ($n % 23 == 0);} $n = 1191216;if (isDivisible($n))echo "Yes" . "\n";elseecho "No" . "\n"; // This code is contributed// by ChitraNayal?>
<script> // JavaScript program to validate above logic // Function to check if the// number is divisible by// 23 or notfunction isDivisible(n){    // While there are at    // least 3 digits    while (Math.floor(n / 100) != 0)    {        // Extracting the last digit        let d = n % 10;                   n = Math.floor(n/10); // Truncating the number           // Adding seven times the last        // digit to the remaining number        n += d * 7;    }       return (n % 23 == 0);} // Driver Codelet n = 1191216;if(isDivisible(n))    document.write("Yes");else    document.write("No");      // This code is contributed by rag2127 </script>
Yes
 
Note that the above program may not make a lot of sense as could simply do n % 23 to check for divisibility. The idea of this program is to validate the concept. Also, this might be an efficient approach if input number is large and given as string. 
Mithun Kumar
ankthon
ukasp
rag2127
divisibility
Number Divisibility
Mathematical
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