Question

I’m trying to complete this Project Euler challenge:

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can
see that the 6th prime is 13.

What is the 10 001st prime number?

My code seem to be right because it works with small numbers, e.g 6th prime is 13.

How can i improve it so that the code will run much more quickly for larger numbers such as 10 001.

Code is below:

#Checks if a number is a prime
def is_prime(n):
    count = 0
    for i in range(2, n):
        if n%i == 0:
            return False
            break
        else:
            count += 1
    if count == n-2:
        return True

#Finds the value for the given nth term     
def term(n):
    x = 0
    count = 0
    while count != n:
        x += 1
        if is_prime(x) == True:
            count += 1
    print x


term(10001)

UPDATE:

Thanks for your responses. I should have been more clear, I am not looking to speed up the interpreter or finding a faster interpreter, because i know my code isn’t great, so i was looking for ways of make my code more efficient.

Answer 1

A few questions to ponder:

• Do you really need to check the division until n-1? How earlier can you stop?
• Apart from 2, do you really need to check the division by all the multiples of two ?
• What about the multiples of 3? 5? Is there a way to extend this idea to all the multiples of previously tested primes?