# Product of all the elements in an array divisible by a given number K

Given an array containing N elements and a number K. The task is to find the product of all such elements of the array which are divisible by K.

Examples:

```Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
K = 3
Output : 810

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
K = 2
Output : 384
```

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

The idea is to traverse the array and check the elements one by one. If an element is divisible by K then multiply that element’s value with the product so far and continue this process while the end of the array is reached.

Below is the implementation of the above approach:

## C++

 `// C++ program to find Product of all the elements ` `// in an array divisible by a given number K ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find Product of all the elements ` `// in an array divisible by a given number K ` `int` `findProduct(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``int` `prod = 1; ` ` `  `    ``// Traverse the array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``// If current element is divisible by k ` `        ``// multiply with product so far ` `        ``if` `(arr[i] % k == 0) { ` `            ``prod *= arr[i]; ` `        ``} ` `    ``} ` ` `  `    ``// Return calculated product ` `    ``return` `prod; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 15, 16, 10, 9, 6, 7, 17 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 3; ` ` `  `    ``cout << findProduct(arr, n, k); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to find Product of all the elements ` `// in an array divisible by a given number K ` ` `  `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `// Function to find Product of all the elements ` `// in an array divisible by a given number K ` `static` `int` `findProduct(``int` `arr[], ``int` `n, ``int` `k) ` `{ ` `    ``int` `prod = ``1``; ` ` `  `    ``// Traverse the array ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` ` `  `        ``// If current element is divisible by k ` `        ``// multiply with product so far ` `        ``if` `(arr[i] % k == ``0``) { ` `            ``prod *= arr[i]; ` `        ``} ` `    ``} ` ` `  `    ``// Return calculated product ` `    ``return` `prod; ` `} ` ` `  `// Driver code ` `    ``public` `static` `void` `main (String[] args) { ` `        ``int` `arr[] = { ``15``, ``16``, ``10``, ``9``, ``6``, ``7``, ``17` `}; ` `    ``int` `n = arr.length; ` `    ``int` `k = ``3``; ` ` `  `    ``System.out.println(findProduct(arr, n, k)); ` `    ``} ` `} ` ` `  ` `  `// This code is contributed by inder_verma.. `

## Python3

 `# Python3 program to find Product of all  ` `# the elements in an array divisible by ` `# a given number K ` ` `  `# Function to find Product of all the elements ` `# in an array divisible by a given number K ` `def` `findProduct(arr, n, k): ` ` `  `    ``prod ``=` `1` ` `  `    ``# Traverse the array ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# If current element is divisible  ` `        ``# by k, multiply with product so far ` `        ``if` `(arr[i] ``%` `k ``=``=` `0``): ` `            ``prod ``*``=` `arr[i] ` ` `  `    ``# Return calculated product ` `    ``return` `prod ` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``arr``=` `[``15``, ``16``, ``10``, ``9``, ``6``, ``7``, ``17` `] ` `    ``n ``=` `len``(arr) ` `    ``k ``=` `3` ` `  `    ``print` `(findProduct(arr, n, k)) ` ` `  `# This code is contributed by ita_c `

## C#

 `// C# program to find Product of all  ` `// the elements in an array divisible ` `// by a given number K ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to find Product of all  ` `// the elements in an array divisible ` `// by a given number K ` `static` `int` `findProduct(``int` `[]arr, ``int` `n, ``int` `k) ` `{ ` `    ``int` `prod = 1; ` ` `  `    ``// Traverse the array ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` ` `  `        ``// If current element is divisible  ` `        ``// by k multiply with product so far ` `        ``if` `(arr[i] % k == 0)  ` `        ``{ ` `            ``prod *= arr[i]; ` `        ``} ` `    ``} ` ` `  `    ``// Return calculated product ` `    ``return` `prod; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 15, 16, 10, 9, 6, 7, 17 }; ` `    ``int` `n = arr.Length; ` `    ``int` `k = 3; ` `     `  `    ``Console.WriteLine(findProduct(arr, n, k)); ` `} ` `} ` ` `  `// This code is contributed by inder_verma `

## PHP

 ` `

Output:

```810
```

Time Complexity: O(N), where N is the number of elements in the array.

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