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def is_prime(x):
 if x < 2:
 return False
 else:
 for n in range(2, x):
 if x % n == 0:
 return False
 else:
 return True 

print is_prime(9) returns True instead of False.

I don't quite understand.

The range (2,9) includes this list: 2,3,4,5,6,7,8

and 9 % 3 == 0,
So how come I do not get False as the answer of that function?

This is because you don't actually loop, as you return True during the first cycle (9 % 2 == 0 is False).

Something like this should solve the problem:

def is_prime(x):
 if x < 2:
 return False
 for n in range(2, x):
 if x % n == 0:
 return False
 return True

You can simplify the logic a good amount by keeping your original loop and not exiting early. You can add your first conditional to your final return:

def is_prime(x):
 for n in range(2, x):
 if x % n == 0:
 return False

 return x > 2

BTW, the Sieve of Erastothenes is a pretty cool method of solving this problem in a much better run time complexity. Here's a link to a brief explanation:

https://math.stackexchange.com/questions/58799/why-in-sieve-of-erastothenes-of-n-number-you-need-to-check-and-cross-out-numbe