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.