Check if a number is divisible by 23 or not

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\equiv1 mod 23.
It can be observed that the smallest n which satisfies this property is 7 as 70\equiv1 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++

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// 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;
}

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Java

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// 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

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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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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Improved By : Mithun Kumar, AnkitRai01, Ita_c