# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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



Output:

55

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