# 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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` 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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