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Sum of the natural numbers (up to N) whose modulo with K yield R
  • Difficulty Level : Expert
  • Last Updated : 20 Feb, 2019

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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

55

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