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

C++Server Side ProgrammingProgramming

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.

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