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 its 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
?>

Javascript

<script>

// Javascript implementation of the above approach

// below function checks whether
// product of digits at even places
// is divisible by K
function productDivisible(n, k)
{
    var product = 1, position = 1;
    while (n > 0) {

        // if position is even
        if (position % 2 == 0)
            product *= n % 10;
        n =parseInt(n / 10);
        position++;
    }

    if (product % k == 0)
        return true;
    return false;
}

// Driver code
var n = 321922;
var k = 3;
if (productDivisible(n, k))
    document.write( "YES");
else
    document.write( "NO");

</script>

Output:

YES

Time Complexity: O(log₁₀n)

Auxiliary Space: O(1)

Method #2:Using string() method:

1. Convert the integer to string then traverse the string and Multiply all even indices by storing it in the product.
2. If the product is divisible by k then return True else False.

Below is the implementation:

C++

// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;

// Function checks whether
// product of digits at even places
// is divisible by K
bool productDivisible(int n, int k)
{
  int product = 1;

  // Converting integer to string
  string num = to_string(n);

  // Traversing the string
  for (int i = 0; i < num.length(); i++) {
    if (i % 2 == 0) {
      product = product * (num[i] - '0');
    }
  }
  if (product % k == 0) {
    return true;
  }
  else {
    return false;
  }
}

// Driver code
int main()
{
  int n = 321922;
  int k = 3;
  if (productDivisible(n, k)) {
    cout << "YES" << endl;
  }
  else {
    cout << "NO" << endl;
  }
}

// This code is contributed by phasing17

Java

// Java implementation of the
// above approach
import java.util.*;

class GFG {

  // Function checks whether
  // product of digits at even places
  // is divisible by K
  static boolean productDivisible(int n, int k)
  {
    int product = 1;

    // Converting integer to string
    String num = String.valueOf(n);

    // Traversing the string
    for (int i = 0; i < num.length(); i++) {
      if (i % 2 == 0) {
        product = product * (num.charAt(i) - '0');
      }
    }
    if (product % k == 0) {
      return true;
    }
    else {
      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");
    }
  }
}

// This code is contributed by phasing17

Python3

# Python3 implementation of the
# above approach

# Function checks whether
# product of digits at even places
# is divisible by K
def productDivisible(n, k):
    product = 1

    # Converting integer to string
    num = str(n)

    # Traversing the string
    for i in range(len(num)):
        if(i % 2 == 0):
            product = product*int(num[i])

    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 vikkycirus

C#

// C# implementation of the
// above approach
using System;
using System.Collections.Generic;

class GFG
{

  // Function checks whether
  // product of digits at even places
  // is divisible by K
  static bool productDivisible(int n, int k)
  {
    int product = 1;

    // Converting integer to string
    string num = Convert.ToString(n);

    // Traversing the string
    for (int i = 0; i < num.Length; i++) {
      if (i % 2 == 0) {
        product = product * (num[i] - '0');
      }
    }
    if (product % k == 0) {
      return true;
    }
    else {
      return false;
    }
  }

  // Driver code
  public static void Main(string[] args)
  {
    int n = 321922;
    int k = 3;
    if (productDivisible(n, k)) {
      Console.WriteLine("YES");
    }
    else {
      Console.WriteLine("NO");
    }
  }
}

// This code is contributed by phasing17

Javascript

<script>

// JavaScript implementation of the
// above approach

// Function checks whether
// product of digits at even places
// is divisible by K
function productDivisible(n, k){
    var product = 1 ;

    // Converting integer to string
    var num = n.toString()

    // Traversing the string
    for(let i = 0 ; i < num.length ; i++){
        if(i % 2 == 0){
            product = product * Number(num[i])
        }
    }
    if (product % k == 0){
        return true
    }
    else{
    return false;
    }
}


// Driver code
var n = 321922
var k = 3
if(productDivisible(n, k)){
    document.write("YES")
}
else{
    document.write("NO")
}


</script>

Output:

YES