找到 13195 的最大素数

Question

I'm trying to solve a problem from the Project Euler archive: https://projecteuler.net/problem=3

The prime factors of 13195 are 5, 7, 13 and 29. What is the largest prime factor of the number 600851475143?

I tried solving for 13195 first. My original idea was to create a list with all the prime numbers lower than the given number. I used the Sieve of Erathostenes algorithm to do this. Then, using a for loop I iterated through all those prime numbers adding the ones that are factors of the given number to a separate list. I sorted the list and printed the largest factor. When I ran the script I got one extra number in the factors list that was not a factor of the given number.

for num = 13915 the output should have been [13, 29, 5, 7] but instead I got [13, 29, 377, 5, 7]

I wasn't able to figure out where 377 came from. I'm aware my solution is not the most efficient one but I tried solving this problem myself. So my questions are:

Where did 377 come from? Can you suggest a more efficient way to solve this problem?

I apologize for any inconveniences as I'm new to problem solving in Python.

primes = []
num = 13195
factors = []
for i in range(1, num + 1):
    #sieve of erathostenes
    if i % 2 != 0 and i % 3 != 0 and i % 5 != 0 and i % 7 != 0 and i != 1:
        primes.append(i)

if num >= 10:
    primes.append(2)
    primes.append(3)
    primes.append(5)
    primes.append(7)

sorted(primes)

for i in range(len(primes)):
    prime = primes[i]
    if num % prime == 0:
        factors.append(primes[i])


print(factors)
factors.sort()
print(factors[len(factors) - 1])

Answer 1

You don't need to generate primes. A direct factorisation of the number will be more efficient:

def primeFactors(N):  # generator for all prime factors
    p = 2
    while p*p<=N:       # only need up to √N (remaining N)
        while N%p == 0: # p is a factor (will be a prime)
            yield p     # return primes as many times as they appear
            N //= p     # remove the prime factor from the number
        p += 1 + p%2    # next potential prime
    if N>1: yield N     # beyond √N whatever remains is a prime factor

output: (take the maximum factor found)

print(max(primeFactors(600851475143))) # 6857

Answer 2

use this to get the factors of a number sorted from largest to smallest

number = 13915
list_of_factors=sorted([int(number/n) for n in range(2,number//2 +1) if number % n ==0 ],reverse=True) 

result is [2783, 1265, 605, 253, 121, 115, 55, 23, 11, 5]