I am looking for prime factors of 2500 with the code below, but my code only prints 2 currently and I am unsure why this is the case.
no = 2500
count = 0
# Finding factors of 2500
for i in range(1,no):
if no%i == 0:
# Now that the factors have been found, the prime factors will be determined
for x in range(1,no):
if i%x==0:
count = count + 1
"""Checking to see if the factor of 2500, itself only has two factor implying it is prime"""
if count == 2:
print i
Thanks
using sieve of eratosthenes to first generate list of primes:
from math import sqrt
def sieve_of_eratosthenes(n):
primes = range(3, n + 1, 2) # primes above 2 must be odd so start at three and increase by 2
for base in xrange(len(primes)):
if primes[base] is None:
continue
if primes[base] >= sqrt(n): # stop at sqrt of n
break
for i in xrange(base + (base + 1) * primes[base], len(primes), primes[base]):
primes[i] = None
primes.insert(0,2)
sieve=filter(None, primes)
return sieve
def prime_factors(sieve,n):
p_f = []
for prime in sieve:
while n % prime == 0:
p_f.append(prime)
n /= prime
if n > 1:
p_f.append(n)
return p_f
sieve = sieve_of_eratosthenes(2500)
print prime_factors(sieve,2500)