# Sum of array elements which are multiples of a given number

• Difficulty Level : Basic
• Last Updated : 04 May, 2021

Given an array arr[] consisting of positive integers and an integer N, the task is to find the sum of all array elements which are multiples of N

Examples:

Input: arr[] = {1, 2, 3, 5, 6}, N = 3
Output: 9
Explanation: From the given array, 3 and 6 are multiples of 3. Therefore, sum = 3 + 6 = 9.

Input: arr[] = {1, 2, 3, 5, 7, 11, 13}, N = 5
Output: 5

Approach: The idea is to traverse the array and for each array element, check if it is a multiple of N or not and add those elements. Follow the steps below to solve the problem:

1. Initialize a variable, say sum, to store the required sum.
2. Traverse the given array and for each array element, perform the following operations.
3. Check whether the array element is a multiple of N or not.
4. If the element is a multiple of N, then add the element to sum.
5. Finally, print the value of sum.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std;`   `// Function to find the sum of array` `// elements which are multiples of N` `void` `mulsum(``int` `arr[], ``int` `n, ``int` `N)` `{`   `    ``// Stores the sum` `    ``int` `sum = 0;`   `    ``// Traverse the given array` `    ``for` `(``int` `i = 0; i < n; i++) {`   `        ``// If current element` `        ``// is a multiple of N` `        ``if` `(arr[i] % N == 0) {` `            ``sum = sum + arr[i];` `        ``}` `    ``}`   `    ``// Print total sum` `    ``cout << sum;` `}`   `// Driver Code` `int` `main()` `{`   `    ``// Given arr[]` `    ``int` `arr[] = { 1, 2, 3, 5, 6 };`   `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``int` `N = 3;`   `    ``mulsum(arr, n, N);`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;` `class` `GFG{`   `// Function to find the sum of array` `// elements which are multiples of N` `static` `void` `mulsum(``int` `arr[], ``int` `n, ``int` `N)` `{`   `    ``// Stores the sum` `    ``int` `sum = ``0``;`   `    ``// Traverse the given array` `    ``for` `(``int` `i = ``0``; i < n; i++)` `    ``{`   `        ``// If current element` `        ``// is a multiple of N` `        ``if` `(arr[i] % N == ``0``) ` `        ``{` `            ``sum = sum + arr[i];` `        ``}` `    ``}`   `    ``// Print total sum` `    ``System.out.println(sum);` `}`     `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    `  `    ``// Given arr[]` `    ``int` `arr[] = { ``1``, ``2``, ``3``, ``5``, ``6` `};` `    ``int` `n = arr.length;` `    ``int` `N = ``3``;` `    ``mulsum(arr, n, N);` `}` `}`   `// This code is contributed by jana_sayantan.`

## Python

 `# Python3 program for the above approach` ` `  `# Function to find the sum of array` `# elements which are multiples of N` `def` `mulsum(arr, n, N):` `     `  `    ``# Stores the sum` `    ``sums ``=` `0` ` `  `    ``# Traverse the array` `    ``for` `i ``in` `range``(``0``, n):` `        ``if` `arr[i] ``%` `N ``=``=` `0``:` `              ``sums ``=` `sums ``+` `arr[i]` ` `  `    ``# Print total sum` `    ``print``(sums)` ` `  `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` ` `  `    ``# Given arr[]` `    ``arr ``=` `[ ``1``, ``2``, ``3``, ``5``, ``6` `]` ` `  `    ``n ``=` `len``(arr)` `    `  `    ``N ``=` `3` ` `  `    ``# Function call` `    ``mulsum(arr, n, N)`

## C#

 `// C# program for the above approach` `using` `System;` `public` `class` `GFG` `{`   `// Function to find the sum of array` `// elements which are multiples of N` `static` `void` `mulsum(``int``[] arr, ``int` `n, ``int` `N)` `{`   `    ``// Stores the sum` `    ``int` `sum = 0;`   `    ``// Traverse the given array` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``{`   `        ``// If current element` `        ``// is a multiple of N` `        ``if` `(arr[i] % N == 0) ` `        ``{` `            ``sum = sum + arr[i];` `        ``}` `    ``}`   `    ``// Print total sum` `    ``Console.Write(sum);` `}`   `// Driver Code` `static` `public` `void` `Main ()` `{` `    ``// Given arr[]` `    ``int``[] arr = { 1, 2, 3, 5, 6 };` `    ``int` `n = arr.Length;` `    ``int` `N = 3;` `    ``mulsum(arr, n, N);` `}` `}`   `// This code is contributed by Dharanendra L V.`

## Javascript

 ``

Output:

`9`

Time Complexity: O(N)
Auxiliary Space: O(1)

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