Sum of the natural numbers (up to N) whose modulo with K yield R

Given three integers N, K, and R. The task is to calculate the sum of all those numbers from 1 to N which yields remainder R upon division by K.
Examples:

Input: N = 20, K = 4, R = 3 
Output: 55 
3, 7, 11, 15 and 19 are the only numbers that give 3 as the remainder on division with 4. 
3 + 7 + 11 + 15 + 19 = 55
Input: N = 15, K = 13, R = 2 
Output: 17 

Approach:

• Initialize sum = 0 and take the modulo of each element from 1 to N with K.
• If the remainder is equal to R, then update sum = sum + i where i is the current number that gave R as the remainder on dividing by K.
• Print the value of sum in the end.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the sum
long long int count(int N, int K, int R)
{
    long long int sum = 0;
    for (int i = 1; i <= N; i++) {

        // If current number gives R as the
        // remainder on dividing by K
        if (i % K == R)

            // Update the sum
            sum += i;
    }

    // Return the sum
    return sum;
}

// Driver code
int main()
{
    int N = 20, K = 4, R = 3;
    cout << count(N, K, R);

    return 0;
}

Java

// Java implementation of the approach
class GfG
{

// Function to return the sum
static long count(int N, int K, int R)
{
    long sum = 0;
    for (int i = 1; i <= N; i++)
    {

        // If current number gives R as the
        // remainder on dividing by K
        if (i % K == R)

            // Update the sum
            sum += i;
    }

    // Return the sum
    return sum;
}

// Driver code
public static void main(String[] args)
{
    int N = 20, K = 4, R = 3;
    System.out.println(count(N, K, R));
}
}

// This code is contributed by
// prerna saini.

Python3

# Python 3 implementation of the approach

# Function to return the sum
def count(N, K, R):
    sum = 0
    for i in range(1, N + 1):

        # If current number gives R as the
        # remainder on dividing by K
        if (i % K == R):

            # Update the sum
            sum += i

    # Return the sum
    return sum

# Driver code
if __name__ == '__main__':
    N = 20
    K = 4
    R = 3
    print(count(N, K, R))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
class GFG
{

// Function to return the sum
static long count(int N, int K, int R)
{
    long sum = 0;
    for (int i = 1; i <= N; i++)
    {

        // If current number gives R as the
        // remainder on dividing by K
        if (i % K == R)

            // Update the sum
            sum += i;
    }

    // Return the sum
    return sum;
}

// Driver code
public static void Main()
{
    int N = 20, K = 4, R = 3;
    Console.Write(count(N, K, R));
}
}

// This code is contributed by
// Akanksha Rai

PHP

<?php
// PHP implementation of the approach

// Function to return the sum
function count1($N, $K, $R)
{
    $sum = 0;
    for ($i = 1; $i <= $N; $i++)
    {

        // If current number gives R as the
        // remainder on dividing by K
        if ($i % $K == $R)

            // Update the sum
            $sum += $i;
    }

    // Return the sum
    return $sum;
}

// Driver code
$N = 20; $K = 4; $R = 3;
echo count1($N, $K, $R);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// Javascript implementation of the approach

    // Function to return the sum
    function count(N , K , R) {
        var sum = 0;
        for (i = 1; i <= N; i++) {

            // If current number gives R as the
            // remainder on dividing by K
            if (i % K == R)

                // Update the sum
                sum += i;
        }

        // Return the sum
        return sum;
    }

    // Driver code

        var N = 20, K = 4, R = 3;
        document.write(count(N, K, R));

// This code contributed by aashish1995

</script>

Output:

55

Time Complexity: O(N)

Auxiliary Space: O(1), since no extra space has been taken.