# Find all divisors of a natural number - Set 2 in C++

In this tutorial, we are going to write a program that finds all the divisors of a natural number. It's a straightforward problem. Let's see the steps to solve it.

• Initialize the number.

• Write a loop that iterates from 1 to the square root of the given number.

• Check whether the given number is divisible by the current number or not.

• If the above condition satisfies, then print the current number and given_number/current_number﻿.

## Example

Let's see the code.

Live Demo

#include <bits/stdc++.h>
using namespace std;
void findDivisors(int n) {
for (int i = 1; i <= sqrt(n); i++) {
if (n % i == 0) {
if (n / i == i) {
cout << i << " ";
}
else {
cout << i << " " << n / i << " ";
}
}
}
cout << endl;
}
int main() {
findDivisors(65);
return 0;
}

## Output

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

1 65 5 13

## Conclusion

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