# 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


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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+c 0 (mod 23)
10(10a+b)+c 0 (mod 23)
10 +c 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 1.
In other words, we have to find an integer such that n such that 10n 1 mod 23.
It can be observed that the smallest n which satisfies this property is 7 as 70 1 mod 23.
Now we can multiply the original equation 10 +c 0 (mod 23)
by 7 and simplify it:
70 +7c 0 (mod 23) +7c 0 (mod 23)
We have found out that if 0 (mod 23) then, +7c 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  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

 

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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