Euler 7

So the other night I was a bit bored and decided to do something to pass the time. I first came across Project Euler a while ago, but had never gone further than problem #1. Boredom is a great motivator and I went through problems #2 thru #9 last night and I decided to post my solutions in search of better ones. Feel free to comment with your suggestions.

Project Euler’s Problem #7 statement is —

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 10001st prime number?

It’s a straightforward problem to solve using brute force, but I went with a mixed approach doing some math instead.

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#!/usr/bin/python
import math

def is_prime(n):
    return not (n < 2 or any(n % x == 0 for x in xrange(2, int(n ** 0.5) + 1)))

currentMax =0
primes = 1
counter = 3

while (primes < 10001):
    if (is_prime(counter)):
        currentMax = counter
        primes+=1
        counter+=2

print currentMax

I initially used a dumb brute-force which ended up running fast enough. But reading about primes, I came across this property by which if a number cannot be divided by any number smaller than sqrt(n), it is a prime. That required significantly less tests to be performed inside is_prime, but then again, it had to perform a square root calculation for every number we need to test. In the end, the algorithm above performed slightly better (0.02s to be exact.)

Updated: Thanks to Eduardo Habkost for pointing out the fact that I had mistakenly pasted the wrong version of the script. I fixed the text to reflect that.