Sum of Multiples

Given two integers n and k, consider all multiples of k that are less than or equal to n. Return the sum of these multiples.

Examples:

Input: n = 5, k = 2
Output: 6
Explanation: The multiples of 2 that do not exceed 5 are 2 and 4. Their sum is 6.

Input: n = 5, k = 3
Output: 3
Explanation: The only multiple of 3 that does not exceed 5 is 3.

[Naive Approach] Check Every Number - O(n) Time and O(1) Space

The idea is to traverse all numbers from 1 to n and check whether each number is divisible by k. If a number is a multiple of k, add it to the answer. After checking every number, return the accumulated sum.

Working of Approach:

• Initialize a variable sum to store the answer.
• Traverse every number from 1 to n.
• For each number, check if it is divisible by k using the modulo operator.
• If it is divisible, add it to sum.
• Return the final value of sum.

#include <iostream>
using namespace std;

int multipleSum(int n, int k)
{

    // Stores the sum of all multiples of k.
    int sum = 0;

    // Traverse all numbers from 1 to n.
    for (int i = 1; i <= n; i++)
    {

        // If i is a multiple of k, add it to the sum.
        if (i % k == 0)
        {
            sum += i;
        }
    }

    // Return the final sum.
    return sum;
}

int main()
{
    int n = 5, k = 3;

    cout << multipleSum(n, k);

    return 0;
}

public class GFG {

    static int multipleSum(int n, int k)
    {

        // Stores the sum of all multiples of k.
        int sum = 0;

        // Traverse all numbers from 1 to n.
        for (int i = 1; i <= n; i++) {

            // If i is a multiple of k, add it to the sum.
            if (i % k == 0) {
                sum += i;
            }
        }

        // Return the final sum.
        return sum;
    }

    public static void main(String[] args)
    {
        int n = 5, k = 3;

        System.out.println(multipleSum(n, k));
    }
}

def multipleSum(n, k):

    # Stores the sum of all multiples of k.
    sum = 0

    # Traverse all numbers from 1 to n.
    for i in range(1, n + 1):

        # If i is a multiple of k, add it to the sum.
        if i % k == 0:
            sum += i

    # Return the final sum.
    return sum


if __name__ == "__main__":
    n, k = 5, 3

    print(multipleSum(n, k))

using System;

class GFG {
    static int multipleSum(int n, int k)
    {
        // Stores the sum of all multiples of k.
        int sum = 0;

        // Traverse all numbers from 1 to n.
        for (int i = 1; i <= n; i++) {
            // If i is a multiple of k, add it to the sum.
            if (i % k == 0) {
                sum += i;
            }
        }

        // Return the final sum.
        return sum;
    }

    static void Main()
    {
        int n = 5, k = 3;

        Console.WriteLine(multipleSum(n, k));
    }
}

function multipleSum(n, k)
{

    // Stores the sum of all multiples of k.
    let sum = 0;

    // Traverse all numbers from 1 to n.
    for (let i = 1; i <= n; i++) {

        // If i is a multiple of k, add it to the sum.
        if (i % k === 0) {
            sum += i;
        }
    }

    // Return the final sum.
    return sum;
}

// Driver Code
let n = 5, k = 3;
console.log(multipleSum(n, k));

3

[Expected Approach] Using Arithmetic Progression Formula - O(1) Time and O(1) Space

The idea is to observe that all multiples of k up to n form an arithmetic progression: k, 2k, 3k, ..., cnt × k, where cnt = n / k. Instead of iterating through all numbers, compute the sum directly using the arithmetic progression formula. To reduce the risk of integer overflow during multiplication, divide either cnt or cnt + 1 by 2 before multiplying.

Working of Approach:

• Compute the number of multiples as cnt = n / k.
• The required sum is k × cnt × (cnt + 1) / 2.
• Since one of cnt or cnt + 1 is always even, divide the even value by 2 first.
• This keeps intermediate multiplication smaller and reduces the risk of integer overflow.
• Return the computed sum.

Let us understand with an example:
Input: n = 5, k = 3

• First, compute the number of multiples of 3 not exceeding 5: cnt = 5 / 3 = 1.
• Since cnt is odd, use the formula cnt × ((cnt + 1) / 2) × k.
• The sum becomes 1 × ((1 + 1) / 2) × 3 = 1 × 1 × 3 = 3.
• The only multiple of 3 that is less than or equal to 5 is 3.
• Hence, the required sum of all multiples is 3.

#include <iostream>
using namespace std;

int multipleSum(int n, int k)
{
    int cnt = n / k;
    int res;

    // Calculate the sum of all multiples of k up to n.
    if (cnt % 2 == 0)
    {
        res = (cnt / 2) * (cnt + 1) * k;
    }
    else
    {
        res = cnt * ((cnt + 1) / 2) * k;
    }

    return res;
}

int main()
{
    int n = 5, k = 3;

    cout << multipleSum(n, k);

    return 0;
}

public class GFG {
    int multipleSum(int n, int k)
    {
        int cnt = n / k;
        int res;

        // Calculate the sum of all multiples of k up to n.
        if (cnt % 2 == 0) {
            res = (cnt / 2) * (cnt + 1) * k;
        }
        else {
            res = cnt * ((cnt + 1) / 2) * k;
        }

        return res;
    }

    public static void main(String[] args)
    {
        GFG obj = new GFG();
        int n = 5, k = 3;

        System.out.println(obj.multipleSum(n, k));
    }
}

def multipleSum(n, k):
    cnt = n // k
    res = 0

    # Calculate the sum of all multiples of k up to n.
    if cnt % 2 == 0:
        res = (cnt // 2) * (cnt + 1) * k
    else:
        res = cnt * ((cnt + 1) // 2) * k

    return res


if __name__ == "__main__":
    n, k = 5, 3

    print(multipleSum(n, k))

using System;

class GFG {
    static int multipleSum(int n, int k)
    {
        int cnt = n / k;
        int res;

        // Calculate the sum of all multiples of k up to n.
        if (cnt % 2 == 0) {
            res = (cnt / 2) * (cnt + 1) * k;
        }
        else {
            res = cnt * ((cnt + 1) / 2) * k;
        }

        return res;
    }

    static void Main()
    {
        int n = 5, k = 3;

        Console.WriteLine(multipleSum(n, k));
    }
}

function multipleSum(n, k)
{
    let cnt = Math.floor(n / k);
    let res;

    // Calculate the sum of all multiples of k up to n.
    if (cnt % 2 === 0) {
        res = (cnt / 2) * (cnt + 1) * k;
    }
    else {
        res = cnt * ((cnt + 1) / 2) * k;
    }

    return res;
}

// Driver Code
let n = 5, k = 3;
console.log(multipleSum(n, k));

3