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Check whether product of digits at even places of a number is divisible by K
  • Last Updated : 21 Nov, 2018

Given a number N, the task is to check whether the product of digits at even places of a number is divisible by K. If it is divisible, output “YES” otherwise output “NO”.

Examples:

Input: N = 5478, K = 5
Output: YES
Since, 5 * 7 = 35, which is divisible by 5

Input: N = 19270, K = 2
Output: NO

Approach:

  1. Find product of digits at even places from right to left.
  2. Then check the divisibility by taking it’s modulo with ‘K’
  3. If modulo gives 0, output YES, otherwise output NO

Below is the implementation of the above approach:

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// below function checks whether
// product of digits at even places
// is divisible by K
bool productDivisible(int n, int k)
{
    int product = 1, position = 1;
    while (n > 0) {
  
        // if position is even
        if (position % 2 == 0)
            product *= n % 10;
        n = n / 10;
        position++;
    }
  
    if (product % k == 0)
        return true;
    return false;
}
  
// Driver code
int main()
{
    int n = 321922;
    int k = 3;
  
    if (productDivisible(n, k))
        cout << "YES";
    else
        cout << "NO";
  
    return 0;
}


Java




// JAVA implementation of the above approach 
class GFG {
// below function checks whether 
// product of digits at even places 
// is divisible by K 
  
    static boolean productDivisible(int n, int k) {
        int product = 1, position = 1;
        while (n > 0) {
  
            // if position is even 
            if (position % 2 == 0) {
                product *= n % 10;
            }
            n = n / 10;
            position++;
        }
  
        if (product % k == 0) {
            return true;
        }
        return false;
    }
  
// Driver code 
    public static void main(String[] args) {
        int n = 321922;
        int k = 3;
  
        if (productDivisible(n, k)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}


Python3




# Python3 implementation of the 
# above approach 
  
# below function checks whether 
# product of digits at even places 
# is divisible by K 
def productDivisible(n, k):
    product = 1
    position = 1
    while n > 0:
          
        # if position is even 
        if position % 2 == 0:
            product *= n % 10
        n = n / 10
        position += 1
    if product % k == 0:
        return True
    return False
  
# Driver code
n = 321922
k = 3
if productDivisible(n, k) == True:
    print("YES")
else:
    print("NO")
  
# This code is contributed 
# by Shrikant13


C#




// C# implementation of the above approach
using System;
  
class GFG
{
// below function checks whether
// product of digits at even places
// is divisible by K
static bool productDivisible(int n, int k)
{
    int product = 1, position = 1;
    while (n > 0)
    {
  
        // if position is even
        if (position % 2 == 0)
            product *= n % 10;
        n = n / 10;
        position++;
    }
  
    if (product % k == 0)
        return true;
    return false;
}
  
// Driver code
public static void Main()
{
    int n = 321922;
    int k = 3;
  
    if (productDivisible(n, k))
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)


PHP




<?php
// PHP implementation of the 
// above approach
  
// Below function checks whether
// product of digits at even places
// is divisible by K
function productDivisible($n, $k)
{
    $product = 1;
    $position = 1;
    while ($n > 0)
    {
  
        // if position is even
        if ($position % 2 == 0)
            $product *= $n % 10;
        $n = (int)($n / 10);
        $position++;
    }
  
    if ($product % $k == 0)
        return true;
    return false;
}
  
// Driver code
$n = 321922;
$k = 3;
  
if (productDivisible($n, $k))
    echo "YES";
else
    echo "NO";
  
// This code is contributed by mits
?>


Output:

YES

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