Sum 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 sum of all such elements which are divisible by K.

Examples:

Input : arr[] = {15, 16, 10, 9, 6, 7, 17}
        K = 3
Output : 30

Input : arr[] = {5, 3, 6, 8, 4, 1, 2, 9}
        K = 2
Output : 20

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

Below is the implementation of the above approach:

C++

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// C++ program to find sum of all the elements
// in an array divisible by a given number K
  
#include <iostream>
using namespace std;
  
// Function to find sum of all the elements
// in an array divisible by a given number K
int findSum(int arr[], int n, int k)
{
    int sum = 0;
  
    // Traverse the array
    for (int i = 0; i < n; i++) {
  
        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }
  
    // Return calculated sum
    return sum;
}
  
// Driver code
int main()
{
    int arr[] = { 15, 16, 10, 9, 6, 7, 17 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
  
    cout << findSum(arr, n, k);
  
    return 0;
}

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Java

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// Java program to find sum of all the elements
// in an array divisible by a given number K
  
import java.io.*;
  
class GFG {
  
// Function to find sum of all the elements
// in an array divisible by a given number K
static int findSum(int arr[], int n, int k)
{
    int sum = 0;
  
    // Traverse the array
    for (int i = 0; i < n; i++) {
  
        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }
  
    // Return calculated sum
    return sum;
}
  
    // 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( findSum(arr, n, k));
    }
}
  
// this code is contributed by anuj_67..

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Python3

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# Python3 program to find sum of 
# all the elements in an array
# divisible by a given number K 
  
# Function to find sum of all 
# the elements in an array 
# divisible by a given number K
def findSum(arr, n, k) :
  
    sum = 0
  
    # Traverse the array 
    for i in range(n) :
  
        # If current element is divisible 
        # by k add it to sum 
        if arr[i] % k == 0 :
  
            sum += arr[i]
  
    # Return calculated sum 
    return sum
  
# Driver code
if __name__ == "__main__" :
  
    arr = [ 15, 16, 10, 9, 6, 7, 17]
    n = len(arr)
    k = 3
  
    print(findSum(arr, n, k))
  
# This code is contributed by ANKITRAI1

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C#

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// C# program to find sum of all the elements
// in an array divisible by a given number K
  
using System;
  
public class GFG{
  
// Function to find sum of all the elements
// in an array divisible by a given number K
static int findSum(int []arr, int n, int k)
{
    int sum = 0;
  
    // Traverse the array
    for (int i = 0; i < n; i++) {
  
        // If current element is divisible by k
        // add it to sum
        if (arr[i] % k == 0) {
            sum += arr[i];
        }
    }
  
    // Return calculated sum
    return sum;
}
  
    // Driver code
      
    static public void Main (){ 
      
    int []arr = { 15, 16, 10, 9, 6, 7, 17 };
    int n = arr.Length;
    int k = 3;
  
    Console.WriteLine( findSum(arr, n, k));
    }
}
//This code is contributed by anuj_67..
     

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PHP

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<?php
// PHP program to find sum of all 
// the elements in an array divisible
// by a given number K
  
// Function to find sum of all 
// the elements in an array 
// divisible by a given number K
function findSum($arr, $n, $k)
{
    $sum = 0;
  
    // Traverse the array
    for ($i = 0; $i < $n; $i++)
    {
  
        // If current element is divisible 
        // by k add it to sum
        if ($arr[$i] % $k == 0) 
        {
            $sum += $arr[$i];
        }
    }
  
    // Return calculated sum
    return $sum;
}
  
// Driver code
$arr = array(15, 16, 10, 9, 6, 7, 17);
$n = sizeof($arr);
$k = 3;
  
echo findSum($arr, $n, $k);
  
// This code is contributed 
// by Akanksha Rai(Abby_akku)

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Output:

30

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



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Improved By : vt_m, AnkitRai01, Akanksha_Rai