Check if a number is divisible by 31 or not

Given a number N, the task is to check whether the number is divisible by 31 or not. 
Examples: 
 

Input: N = 1922 
Output: Yes 
Explanation: 
31 * 62 = 1922
Input: N = 2722400 
Output: No 
 

Approach: The divisibility test of 31 is: 
 

1. Extract the last digit.
2. Subtract 3 * last digit from the remaining number obtained after removing the last digit.
3. Repeat the above steps until a two-digit number, or zero, is obtained.
4. If the two-digit number is divisible by 31, or it is 0, then the original number is also divisible by 31.

For example: 
 

If N = 49507

Step 1:
  N = 49507
  Last digit = 7
  Remaining number = 4950
  Subtracting 3 times last digit
  Resultant number = 4950 - 3*7 = 4929

Step 2:
  N = 4929
  Last digit = 9
  Remaining number = 492
  Subtracting 3 times last digit
  Resultant number = 492 - 3*9 = 465

Step 3:
  N = 465
  Last digit = 5
  Remaining number = 46
  Subtracting 3 times last digit
  Resultant number = 46 - 3*5 = 31

Step 4:
  N = 31
  Since N is a two-digit number,
  and 31 is divisible by 31

Therefore N = 49507 is also divisible by 31

Below is the implementation of the above approach: 
 

// C++ program to check whether a number
// is divisible by 31 or not
#include<bits/stdc++.h>
#include<stdlib.h>

using namespace std;

// Function to check if the number is divisible by 31 or not 
bool isDivisible(int n) 
{
    int d;
    
    // While there are at least two digits 
    while (n / 100) 
    {

        // Extracting the last 
        d = n % 10;

        // Truncating the number 
        n /= 10;

        // Subtracting three times the last 
        // digit to the remaining number 
        n = abs(n-(d * 3));
    }
    
    // Finally return if the two-digit
    // number is divisible by 31 or not
    return (n % 31 == 0) ;
}

// Driver Code 
int main() 
{
    int N = 1922;

    if (isDivisible(N)) 
        cout<<"Yes"<<endl ;
    else
        cout<<"No"<<endl ;
    
    return 0;     
} 

// This code is contributed by ANKITKUMAR34

// Java program to check whether a number
// is divisible by 31 or not
import java.util.*;

class GFG{
 
// Function to check if the number is divisible by 31 or not 
static boolean isDivisible(int n) 
{
    int d;
     
    // While there are at least two digits 
    while ((n / 100) > 0) 
    {
 
        // Extracting the last 
        d = n % 10;
 
        // Truncating the number 
        n /= 10;
 
        // Subtracting three times the last 
        // digit to the remaining number 
        n = Math.abs(n - (d * 3));
    }
     
    // Finally return if the two-digit
    // number is divisible by 31 or not
    return (n % 31 == 0) ;
}
 
// Driver Code 
public static void main(String[] args) 
{
    int N = 1922;
 
    if (isDivisible(N)) 
        System.out.print("Yes");
    else
        System.out.print("No");
}     
} 

// This code is contributed by PrinciRaj1992

# Python program to check whether a number
# is divisible by 31 or not

# Function to check if the number is 
# divisible by 31 or not 
def isDivisible(n) : 

    # While there are at least two digits 
    while n // 100 : 

        # Extracting the last 
        d = n % 10

        # Truncating the number 
        n //= 10

        # Subtracting three times the last 
        # digit to the remaining number 
        n = abs(n-(d * 3))

    # Finally return if the two-digit
    # number is divisible by 31 or not
    return (n % 31 == 0) 

# Driver Code 
if __name__ == "__main__" : 

    n = 1922

    if (isDivisible(n)) : 
        print("Yes") 
    else : 
        print("No") 

// C# program to check whether a number
// is divisible by 31 or not
using System;

class GFG{
  
// Function to check if the number is divisible by 31 or not 
static bool isDivisible(int n) 
{
    int d;
      
    // While there are at least two digits 
    while ((n / 100) > 0) 
    {
  
        // Extracting the last 
        d = n % 10;
  
        // Truncating the number 
        n /= 10;
  
        // Subtracting three times the last 
        // digit to the remaining number 
        n = Math.Abs(n - (d * 3));
    }
      
    // Finally return if the two-digit
    // number is divisible by 31 or not
    return (n % 31 == 0) ;
}
  
// Driver Code 
public static void Main(String[] args) 
{
    int N = 1922;
  
    if (isDivisible(N)) 
        Console.Write("Yes");
    else
        Console.Write("No");
}     
}

// This code is contributed by Rajput-Ji

<script>
// Javascript program to check whether a number
// is divisible by 31 or not
  
// Function to check if the number is divisible by 31 or not 
function isDivisible(n) 
{
    let d;
       
    // While there are at least two digits 
    while (Math.floor(n / 100) > 0) 
    {
   
        // Extracting the last 
        d = n % 10;
   
        // Truncating the number 
        n = Math.floor(n / 10);
   
        // Subtracting three times the last 
        // digit to the remaining number 
        n = Math.abs(n - (d * 3));
    }
       
    // Finally return if the two-digit
    // number is divisible by 31 or not
    return (n % 31 == 0) ;
}

// Driver Code
    
    let N = 1922;
    
    if (isDivisible(N) != 0) 
        document.write("Yes") ;
    else
        document.write("No");

// This code is contributed by sanjoy_62.
</script>

Yes

Time Complexity: O(log₁₀N)

Auxiliary Space: O(1)