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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.
#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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