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

Given an array arr[n] with n number of integers and another integer k, the task is to find the product all the elements of arr[] which are divisible by k.

To solve the problem we have to iterate every element of the array and find whether it is completely divisible by the number k and then product all the elements and store it into a variable. Like we have an array arr[] = {1, 2, 3, 4, 5, 6 } and assuming we have k = 2 so the numbers in the array which are divisible by 2 are 2, 4, 6 and their product will be equal to 48.

So, let's see the example how we want our answers as per the input

Input

arr[] = {10, 11, 55, 2, 6, 7}
K = 11

Output

Explanation − the numbers divisible by 11 are 11 and 55 only their product is 605

Input

arr[] = {9, 8, 7, 6, 3}
K = 3

Output

Approach used below is as follows to solve the problem

Iterate the whole array till the very end of an array.
Look for every integer which is divisible by K.
Product every element divisible by K.
Return the product.
Print the result.

Algorithm

Start
Step 1→ declare function to find all the numbers divisible by number K
   int product(int arr[], int size, int k)
      declare int prod = 1
      Loop For int i = 0 and i < size and i++
      IF (arr[i] % k == 0)
         Set prod *= arr[i]
      End
      End
      return prod
Step 2→ In main()
   Declare int arr[] = {2, 3, 4, 5, 6 }
   Declare int size = sizeof(arr) / sizeof(arr[0])
   Set int k = 2
   Call product(arr, size, k)
Stop

Example

Live Demo

#include <iostream>
using namespace std;
//function to find elements in an array divisible by k
int product(int arr[], int size, int k){
   int prod = 1;
   for (int i = 0; i < size; i++){
      if (arr[i] % k == 0){
         prod *= arr[i];
      }
   }
   return prod;
}
int main(){
   int arr[] = {2, 3, 4, 5, 6 };
   int size = sizeof(arr) / sizeof(arr[0]);
   int k = 2;
   cout<<"product of elements are : "<<product(arr, size, k);
   return 0;
}