Sum of multiples of a number up to N

Given a number a and limit N. Find the sum of multiple of a upto N.

Examples :

Input : a = 4, N = 23
Output : sum = 60
[Multiples : 4, 8, 12, 16, 20]

Input :a = 7, N = 49
Output :sum = 196
[Multiples: 7, 14, 21, 28, 35, 42, 49]

The basic idea is to iterate from i = a to i = n, i++ and check whether i % a == 0 or not.If zero then add i to sum(initially sum = 0).Thus we will get the sum.It will take O(n) time.
We can modify the loop as i = a, i <= n, i = i + a to reduce the number of iterations.But it will also take O(m) time if there is m multiples of a.
To get the result in O(1) time we can use the formula of summation of n natural numbers.For the above example, 
a = 4 and N = 23, number of multiples of a, m = N/a(integer division). The multiples are 4, 8, 12, 16, 20. 
We can write it as 4 X [1, 2, 3, 4, 5]. So we can get the sum of multiples as:

 sum = a * (Summation of 1 to m [natural numbers from 1 to m]) 
 sum = 4 * (m*(m+1) / 2)
 sum = 4 * (5*6 / 2) = 4 * 15 = 60 
 

C++

// C++ program to find sum of multiples of a number
// up to N efficiently
#include <iostream>
using namespace std;

// Function for calculating sum of multiples of
// a upto N
int calculate_sum(int a, int N)
{
    // Number of multiples
    int m = N / a;

    // sum of first m natural numbers
    int sum = m * (m + 1) / 2;

    // sum of multiples
    int ans = a * sum;

    return ans;
}

// Driver code
int main()
{
    int a = 7, N = 49;
    cout << "Sum of multiples of "
         << a << " up to " << N << " = "
         << calculate_sum(a, N) << endl;
    return 0;
}

Java

// Java program to find sum of multiples
// of a number up to N efficiently

class GFG {

// Function for calculating sum
// of multiples of a upto N
static int calculate_sum(int a, int N) {

    // Number of multiples
    int m = N / a;

    // sum of first m natural numbers
    int sum = m * (m + 1) / 2;

    // sum of multiples
    int ans = a * sum;

    return ans;
}

// Driver code
public static void main(String[] args) {

    int a = 7, N = 49;
    System.out.println("Sum of multiples of " + a +
                       " up to " + N + " = " +
                               calculate_sum(a, N));
}
}

// This code is contributed by Anant Agarwal.

Python3

"""Python program to find sum of
multiples of a number up to N"""

# Calculates sum of multiples of
# a number upto N
def calculate_sum(a, N):

    # Number of multiples
    m = N / a

    # sum of first m natural numbers
    sum = m * (m + 1) / 2

    # sum of multiples
    ans = a * sum

    print("Sum of multiples of ", a,
          " up to ", N, " = ", ans)

# Driver Code
calculate_sum(7, 49)

# This code is contributed by Abhishek Agrawal.

C#

// C# program to find sum of multiples
// of a number up to N efficiently
using System;

class GFG {

    // Function for calculating sum
    // of multiples of a upto N
    static int calculate_sum(int a, int N)
    {

        // Number of multiples
        int m = N / a;

        // sum of first m natural numbers
        int sum = m * (m + 1) / 2;

        // sum of multiples
        int ans = a * sum;

        return ans;
    }

    // Driver code
    public static void Main()
    {

        int a = 7, N = 49;
        Console.WriteLine("Sum of multiples of " + a +
         " up to " + N + " = " + calculate_sum(a, N));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to find sum
// of multiples of a number
// up to N efficiently

// Function for calculating sum
// of multiples of a upto N
function calculate_sum($a, $N)
{
    // Number of multiples
    $m = $N / $a;

    // sum of first m
    // natural numbers
    $sum = $m * ($m + 1) / 2;

    // sum of multiples
    $ans = $a * $sum;

    return $ans;
}

// Driver code
$a = 7;
$N = 49;
echo "Sum of multiples of ". $a ,
         " up to " . $N . " = " .
          calculate_sum($a, $N) ;

// This code is contributed by Sam007
?>

Javascript

<script>

// Javascript program to find sum
// of multiples of a number
// up to N efficiently

// Function for calculating sum
// of multiples of a upto N
function calculate_sum(a, N)
{

    // Number of multiples
    m = N / a;

    // Sum of first m
    // natural numbers
    sum = m * (m + 1) / 2;

    // Sum of multiples
    ans = a * sum;

    return ans;
}

// Driver code
let a = 7;
let N = 49;

document.write("Sum of multiples of "+ a +
               " up to " + N + " = " +
               calculate_sum(a, N));

// This code is contributed by mohan1240760

</script>

Output : 

Sum of multiples of 7 upto 49 = 196

Time complexity: O(1)
Auxiliary space: O(1)

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