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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.
#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.
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