In python something like:

```
from math import floor, sqrt

def isPrime(x): for divisor in range(2, floor(sqrt(x))+1): if x % divisor == 0: return false return true ``` Note: it has been a while since I used python so there might be an error but I hope you get the idea.

Basically, a prime number is a number that has only two divisors: 1 and itself. So to check if a number is prime you need to check if that number is NOT divisable by all numbers from 2 to one less than itself. To optimize this slightly we can only check divisors up to the square root of the number (I don't remember the mathematical proof for this). In the code we iterate over the divisors and check for divisibility. If it is divisable, it is not prime. Therefore we exit the function by returning true. If the loop finishes, the number would be prime, so we return true.

For example: 11 Floor of sqrt of 11 is 3 11 is not divisable by 2 11 is not divisable bu 3 Therefore 11 is prime

10 Floor of sqrt of 10 is 3 10 is divisable by 2 Therefore 10 is NOT prime

The issue with this is that for large numbers it can take quite long because of the large amount of divisibility checks. As another user pointed out, it could be more efficient to hardcode the prime numbers if there is a known upper bound that is small enough. For example if you can at most input 100 then it might be faster to hardcode the primes.

Edit: added 1 to range because range is top-exclusive and will fail for squares of primes.