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#
# 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())