# 
# Solution to Project Euler problem 3
# Copyright (c) Project Nayuki. All rights reserved.
# 
# https://www.nayuki.io/page/project-euler-solutions
# https://github.com/nayuki/Project-Euler-solutions
# 

import eulerlib


# By the fundamental theorem of arithmetic, every integer n > 1 has a unique factorization as a product of prime numbers.
# In other words, the theorem says that n = p_0 * p_1 * ... * p_{m-1}, where each p_i > 1 is prime but not necessarily unique.
# Now if we take the number n and repeatedly divide out its smallest factor (which must also be prime), then the last
# factor that we divide out must be the largest prime factor of n. For reference, 600851475143 = 71 * 839 * 1471 * 6857.
def compute():
	n = 600851475143
	while True:
		p = smallest_prime_factor(n)
		if p < n:
			n //= p
		else:
			return str(n)


# Returns the smallest factor of n, which is in the range [2, n]. The result is always prime.
def smallest_prime_factor(n):
	assert n >= 2
	for i in range(2, eulerlib.sqrt(n) + 1):
		if n % i == 0:
			return i
	return n  # n itself is prime


if __name__ == "__main__":
	print(compute())