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

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 <iostream>` `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[0]);` ` ` `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

`<?php` `// PHP 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` `function` `findProduct(&` `$arr` `, ` `$n` `, ` `$k` `)` `{` ` ` `$prod` `= 1;` ` ` `// Traverse the array` ` ` `for` `(` `$i` `= 0; ` `$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` `$prod` `;` `}` `// Driver code` `$arr` `= ` `array` `(15, 16, 10, 9, 6, 7, 17 );` `$n` `= sizeof(` `$arr` `);` `$k` `= 3;` `echo` `(findProduct(` `$arr` `, ` `$n` `, ` `$k` `));` `// This code is contributed` `// by Shivi_Aggarwal` `?>` |

## Javascript

`<script>` `// Function to find Product of all the elements` `// in an array divisible by a given number K` `function` `findProduct( arr, n, k)` `{` ` ` `var` `prod = 1;` ` ` `// Traverse the array` ` ` `for` `(` `var` `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;` `}` `var` `arr = [15, 16, 10, 9, 6, 7, 17 ];` ` ` ` ` `document.write(findProduct(arr, 7, 3));` `</script>` |

**Output:**

810

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

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