# Number of pairs from the first N natural numbers whose sum is divisible by K in C++

C++Server Side ProgrammingProgramming

In this tutorial, we are going to write a program that counts the pairs whose sum is divisible by K.

Let's see the steps to solve the problem.

• Initialise the N and K.

• Generate the natural numbers till N and store them in an array.

• Compute the sum of every pair.

• If the pair sum is divisible by K, then increment the count.

## Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
int get2PowersCount(vector<int> arr, int N, int K) {
int count = 0;
for (int i = 0; i < N; i++) {
for (int j = i + 1; j < N; j++) {
int sum = arr[i] + arr[j];
if (sum % K == 0) {
count++;
}
}
}
return count;
}
int main() {
vector<int> arr;
int N = 10, K = 5;
for (int i = 1; i <= N; i++) {
arr.push_back(i);
}
cout << get2PowersCount(arr, N, K) << endl;
return 0;
}

## Output

If you run the above code, then you will get the following result.

9

## Conclusion

If you have any queries in the tutorial, mention them in the comment section.