def get_prime_factors(number):
# root = math.floor(math.sqrt(number))
primes_up_to_number = gen_primes(number)
prime_factors = []
for prime in primes_up_to_number:
if number % prime == 0:
prime_factors.append(prime)
return sorted(prime_factors)
def sum_of_primes():
N = 2000000
t = 2
sum = 0
g = gen_primes()
while t < N:
t = g.next()
sum += t
if t < 100 or t > 1900000:
print t, sum
return sum
def solution():
sets = []
for prime in gen_primes():
if prime == 2:
continue
for s in sets:
ispp = all(prime_pair(element, prime) for element in s)
if ispp:
sets.append(s|{prime})
if len(s) + 1 >= 4:
print s | {prime}
if len(s) + 1 == 5:
return sum(s) + prime
sets.append({prime})
return None
""" Project Euler Problem 7 ======================= By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10001st prime number? """ from utils import gen_primes # Listing primes less than 200,000 yields more than enough. print gen_primes(200000)[10000]
def candidates():
for p in gen_primes():
digits = digitize(p)
if len(digits) > len(set(digits)):
yield digits
from utils import gen_primes print(sum(gen_primes(2 * (10**6))))
"""
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13,
we can see that the 6th prime is 13.
What is the 10 001st prime number?
"""
from utils import gen_primes
LIMIT = 10001
prime_gen = gen_primes()
c = 0
while c != LIMIT:
prime = prime_gen.next()
c += 1
print prime
The number 3797 has an interesting property. Being prime itself, it is
possible to continuously remove digits from left to right, and remain
prime at each stage: 3797, 797, 97, and 7. Similarly we can work from
right to left: 3797, 379, 37, and 3.
Find the sum of the only eleven primes that are both truncatable from left
to right and right to left.
NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.
"""
from utils import gen_primes
from collections import defaultdict
primes = gen_primes(1000000)
even = set(str(24680))
primes_by_length = defaultdict(list)
trunc_by_length = defaultdict(list)
for p in primes:
primes_by_length[len(str(p))].append(p)
# truncatable prime can never contain 1,4,6,8,9
if str(p)[0] in (1,4,6,8,9):
continue
if len(str(p))==1:
trunc_by_length[1].append(p)
else:
len_p = len(str(p))
from utils import gen_primes
prime_list = []
limit = 10000
while len(prime_list) < 10001:
prime_list = gen_primes(limit)
limit *= 10
print(prime_list[10000])
def test_gen_primes_start(self):
gp = gen_primes(6)
self.assertEqual([7,11,13,17,19], [gp.next() for i in range(5)])
def test_gen_primes(self):
gp = gen_primes()
self.assertEqual([2,3,5,7,11], [gp.next() for i in range(5)])
""" Project Euler Problem 10 ======================== The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. Find the sum of all the primes below two million. """ from utils import gen_primes N = 2000000 print sum(gen_primes(N-1))
# Problem 3:
# The prime factors of 13195 are 5, 7, 13 and 29.
# What is the largest prime factor of the number 600851475143?
import utils
number = 600851475143
counter = 600851475143
primes = []
for x in utils.gen_primes():
# Break out of the loop if the square of the number
# if larger than the original value being checked against.
if (x * x) > counter:
break
elif number % x == 0:
primes.append(x)
number = number / x
# Return the largest prime factor that was found.
print primes[-1]
"""
Project Euler Problem 3
=======================
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143?
"""
from math import sqrt
from utils import gen_primes
N = 600851475143
max_prime = N
candidates = gen_primes(sqrt(N))
candidates.reverse()
for p in candidates:
if N%p==0:
max_prime = p
break
print max_prime
