Check whether product of digits at even places of a number is divisible by K

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++

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// 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;
}

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Java














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// 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"); 


        } 


    } 


} 



















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Python3














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# 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 



















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C#














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// 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) 



















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PHP














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<?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 


?> 



















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Output:


YES







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