Brute force

$i$ is a factor of $n$ if $n\ \equiv\ 0\ [i]$, in our case it is enough to iterate over each number. When a factor is found, we simply divide $n$ by that factor and continue as long as $n$ is greater than one. When $n$ is equal to one, we simply return the current factor which is also the largest.

From solution1.py:

def largest_prime_factor(n=600851475143):
    res = 2
    while n != 1:
        if n % res == 0:
            n //= res
        else:
            res += 1

    return res