First triangular number whose number of divisors exceeds N in C++

In this tutorial, we are going to find a triangular number whose number of divisors are greater than n.

If the sum of natural numbers at any point less than or equal to n is equal to the given number, then the given number is a triangular number.

We have seen what triangular number is. Let's see the steps to solve the problem.

• Initialize the number

• Write a loop until we find the number that satisfies the given conditions.

• Check whether the number is triangular or not.

• Check whether the number has more than n divisors or not.

• If the above two conditions are satisfied then print the number and break the loop.

Example

Let's see the code.

 Live Demo

#include <bits/stdc++.h>
using namespace std;
bool isTriangular(int n) {
   if (n < 0) {
      return false;
   }
   int sum = 0;
   for (int i = 1; sum <= n; i++) {
      sum += i;
      if (sum == n) {
         return true;
      }
   }
   return false;
}
int divisiorsCount(int n) {
   int count = 0;
   for (int i = 1; i <= n; i++) {
      if (n % i == 0) {
         count += 1;
      }
   }
   return count;
}
int main() {
   int n = 2, i = 1;
   while (true) {
      if (isTriangular(i) && divisiorsCount(i) > 2) {
         cout << i << endl;
         break;
      }
      i += 1;
   }
   return 0;
}

Output

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

6

Conclusion

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