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Check if the given number is divisible by 71 or not

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  • Last Updated : 07 Aug, 2022

Given a number N, the task is to check whether the number is divisible by 71 or not. 

Examples:  

Input: N = 25411681 
Output: yes 
Explanation: 
71 * 357911 = 25411681

Input: N = 5041 
Output: yes 
Explanation: 
71 * 71 = 5041  

Approach: The divisibility test of 71 is:  

  1. Extract the last digit.
  2. Subtract 7 * 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 71, or it is 0, then the original number is also divisible by 71.

For example:  

If N = 5041

Step 1:
  N = 5041
  Last digit = 1
  Remaining number = 504
  Subtracting 7 times last digit
  Resultant number = 504 - 7*1 = 497

Step 2:
  N = 497
  Last digit = 7
  Remaining number = 49
  Subtracting 7 times last digit
  Resultant number = 49 - 7*7 = 0

Step 3:
  N = 0
  Since N is a two-digit number,
  and 0 is divisible by 71

Therefore N = 5041 is also divisible by 71

Below is the implementation of the above approach:  

C++




// C++ program to check whether a number
// is divisible by 71 or not
#include<bits/stdc++.h>
#include<stdlib.h>
 
using namespace std;
 
// Function to check if the number is divisible by 71 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 seven times the last
        // digit to the remaining number
        n = abs(n - (d * 7));
    }
    // Finally return if the two-digit
    // number is divisible by 71 or not
    return (n % 71 == 0) ;
}
 
// Driver Code
int main() {
    int N = 5041;
 
    if (isDivisible(N))
        cout << "Yes" << endl ;
    else
        cout << "No" << endl ;
     
    return 0;    
}
 
// This code is contributed by ANKITKUMAR34

Java




// Java program to check whether a number
// is divisible by 71 or not
import java.util.*;
 
class GFG{
 
// Function to check if the number is divisible by 71 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 seven times the last
            // digit to the remaining number
            n = Math.abs(n - (d * 7));
        }
 
        // Finally return if the two-digit
        // number is divisible by 71 or not
        return (n % 71 == 0) ;
    }
     
    // Driver Code
    public static void main(String args[]){
        int N = 5041;
     
        if (isDivisible(N))
            System.out.println("Yes") ;
        else
            System.out.println("No");
    }
}
 
// This code is contributed by AbhiThakur

Python 3




# Python program to check whether a number
# is divisible by 71 or not
 
# Function to check if the number is
# divisible by 71 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 seven times the last
        # digit to the remaining number
        n = abs(n-(d * 7))
 
    # Finally return if the two-digit
    # number is divisible by 71 or not
    return (n % 71 == 0)
 
# Driver Code
if __name__ == "__main__" :
     
    N = 5041
 
    if (isDivisible(N)) :
        print("Yes")
    else :
        print("No")

C#




// C# program to check whether a number
// is divisible by 71 or not
using System;
         
class GFG
{
     
// Function to check if the number is divisible by 71 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 fourteen times the last
        // digit to the remaining number
        n = Math.Abs(n - (d * 7));
    }
     
    // Finally return if the two-digit
    // number is divisible by 71 or not
    return (n % 71 == 0);
}
     
// Driver Code
public static void Main()
{
    int N = 5041;
     
    if (isDivisible(N))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by mohit kumar 29.

Javascript




<script>
// Javascript program to check whether a number
// is divisible by 71 or not
 
// Function to check if the number is divisible by 71 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 seven times the last
            // digit to the remaining number
            n = Math.abs(n - (d * 7));
        }
   
        // Finally return if the two-digit
        // number is divisible by 71 or not
        return (n % 71 == 0) ;
}
 
// Driver Code
let N = 5041;
if (isDivisible(N))
    document.write("Yes") ;
else
    document.write("No");
 
 
 
// This code is contributed by patel2127
</script>

Output: 

Yes

 

Time Complexity: O(log10N)

Auxiliary Space: O(1)


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