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Check if a number is divisible by 23 or not

  • Last Updated : 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 \overline{a b c}  be any number such that \overline{a b c}  =100a+10b+c . 
Now assume that \overline{a b c}  is divisible by 23. Then 
\overline{a b c}\equiv  0 (mod 23) 
100a+10b+c\equiv  0 (mod 23) 
10(10a+b)+c\equiv  0 (mod 23) 
10\overline{a b}  +c\equiv  0 (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 \overline{a b}  1. 
In other words, we have to find an integer such that n such that 10n\equiv  1 mod 23. 
It can be observed that the smallest n which satisfies this property is 7 as 70\equiv  1 mod 23. 
Now we can multiply the original equation 10\overline{a b}  +c\equiv  0 (mod 23) 
by 7 and simplify it: 
70\overline{a b}  +7c\equiv  0 (mod 23) 
\overline{a b}  +7c\equiv  0 (mod 23) 
We have found out that if \overline{a b c}\equiv  0 (mod 23) then, 
\overline{a b}  +7c\equiv  0 (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++




// CPP program to validate above logic
#include <iostream>
using namespace std;
 
// Function to check if the number is
// divisible by 23 or not
bool 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




// Java program to validate above logic
class GFG
{
     
// Function to check if the
// number is divisible by
// 23 or not
static 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 Code
public 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




# Python 3 program to validate above logic
 
# Function to check if the number is
# divisible by 23 or not
def 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 Code
if __name__ == "__main__" :
 
    n = 1191216
 
    # function calling
    if (isDivisible(n)) :
        print("Yes")
 
    else :
        print("No")
 
# This code is contributed by ANKITRAI1

C#




// C# program to validate
// above logic
class GFG
{
     
// Function to check if the
// number is divisible by
// 23 or not
static 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 Code
public 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
// PHP program to validate above logic
 
// Function to check if the number
// is divisible by 23 or not
function 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";
else
echo "No" . "\n";
 
// This code is contributed
// by ChitraNayal
?>

Javascript




<script>
 
// JavaScript program to validate above logic
 
// Function to check if the
// number is divisible by
// 23 or not
function 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 Code
let n = 1191216;
if(isDivisible(n))
    document.write("Yes");
else
    document.write("No");
     
 
// This code is contributed by rag2127
 
</script>
Output: 
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.
 

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