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Product of all the elements in an array divisible by a given number K
• Difficulty Level : Basic
• Last Updated : 12 May, 2021

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 ``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

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## Javascript

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Output:
`810`

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

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