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In this problem, we are given three numbers N, K and R. Our task is to create a program to find the Sum of the natural numbers (up to N) whose modulo with K yield R.

We will add all the numbers less than N that satisfy the following condition, i%K == R.

**Let’s take an example to understand the problem,**

**Input**

N = 14, K = 4, R = 1

**Output**

28

**Explanation** − All the numbers less than N, that given 1 as remainder when divided by 4 are 1, 5, 9, 13.

To solve this problem, we will loop from R to N, and increment the value by K. So, that we will get eveny number that satisfies the given condition. And add them to the sum.

Here, we could have used to normal loop i.e. with 1 as an interval. But we have used this before it will consume less time.

Program to illustrate the solution,

#include <iostream> using namespace std; int CalcSumofRem(int N, int K, int R){ int sum = 0; for (int i = R; i <= N; i+= K) { if (i % K == R) sum += i; } return sum; } int main(){ int N = 14, K = 4, R = 1; cout<<"Sum of natural numbers (up to "<<N<<") whose modulo with "<<K<<" yields "<<R<<" is "<<CalcSumofRem(N, K, R); return 0; }

Sum of natural numbers (up to 14) whose modulo with 4 yields 1 is 28

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