10001번째 소수. 체크로직을 이용.

primeOf10001stC.py

import time 
start = time.time()

def is_prime(v):
    i = 2
    while True:
        if (v % i) == 0:
            return False
        elif (i * i) > v:
            return True
        i += 1

count_prime = 1
number = 3
while count_prime < 10001:
    if (is_prime(number)):
        count_prime += 1
    number += 1
print number - 1

end = time.time() - start   
print str(end) + ' sec'