def num_divisors(num):
primes = utils.sieve_of_eratosthenes(int(math.ceil(math.sqrt(num))))
factors = []
prime_counter = 0
while num != 1:
if prime_counter > len(primes) - 1:
if num % 2 != 0 and num > 3 and utils.is_prime(num, 10):
factors.append(num)
break
else:
print("fail")
break
if num % primes[prime_counter] == 0:
num /= primes[prime_counter]
factors.append(primes[prime_counter])
else:
prime_counter += 1
c = Counter(factors)
product = 1
for key, value in c.iteritems():
value += 1
product *= value
return product
def prime_factors(num):
primes = utils.sieve_of_eratosthenes(int(math.ceil(math.sqrt(num))))
factors = []
prime_counter = 0
while num != 1:
if num % primes[prime_counter] == 0:
num /= primes[prime_counter]
factors.append(primes[prime_counter])
else:
prime_counter += 1
return factors[len(factors) - 1]
def prime_factors(num):
primes = utils.sieve_of_eratosthenes(int(math.ceil(math.sqrt(num))) + 1)
factors = set()
prime_counter = 0
while num != 1 and prime_counter < len(primes):
if num % primes[prime_counter] == 0:
num /= primes[prime_counter]
factors.add(primes[prime_counter])
else:
prime_counter += 1
if num != 1:
factors.add(int(num))
return factors
def equal_gap(one, two, three):
return two - one == three - two
def check_perms(n):
perms = [int(''.join(p)) for p in itertools.permutations(str(n), 4)]
perms = [perm for perm in perms if utils.is_prime(perm) and perm >= 1000]
for i in range(0, len(perms)):
for j in range(i + 1, len(perms)):
for h in range(j + 1, len(perms)):
if (perms[i] != perms[j] != perms[h] and
equal_gap(perms[i], perms[j], perms[h])):
return (str(perms[i]) + ', ' +
str(perms[j]) + ', ' + str(perms[h]))
primes = utils.sieve_of_eratosthenes(10000)
lower_index = 0
upper_index = len(primes)
for i in range(0, len(primes)):
if primes[i] > 1000:
lower_index = i - 1
break
for i in range(lower_index, upper_index):
check = check_perms(primes[i])
if check is not None:
print(check)
#
# Solution to Project Euler Problem 37
#
import utils
def truncatable(num):
num = str(num)
temp_num = str(num)
for i in range(0, len(num) - 1):
temp_num = temp_num[1:]
if not utils.is_prime(int(temp_num)):
return False
for i in range(0, len(num) - 1):
num = num[:-1]
if not utils.is_prime(int(num)):
return False
return True
primes_so_far = []
for prime in utils.sieve_of_eratosthenes(1000000)[4:]:
if truncatable(prime):
primes_so_far.append(prime)
print(sum(primes_so_far))
# The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
#
# Find the sum of all the primes below two million.
import time
from utils import sieve_of_eratosthenes
if __name__ == '__main__':
t0 = time.time()
results = sieve_of_eratosthenes(2000000)
runtime = time.time() - t0
print 'Runtime = {}'.format(runtime)
print 'Result = {}'.format(sum(results))
#
# Solution to Project Euler Problem 50
#
import utils
prime_list = [x for x in utils.sieve_of_eratosthenes(1000000) if x < 100000]
max_ = 0
for i in range(0, len(prime_list)):
for j in range(i + 1, len(prime_list)):
if sum(prime_list[i:j]) > 1000000:
break
if utils.is_prime(sum(prime_list[i:j])) and j - i > max_:
max_ = j - i
print(str(sum(prime_list[i:j])) + ', ' + str(max_))
#
# where |n| is the modulus/absolute value of n
# e.g. |11| = 11 and |-4| = 4
# Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number
# of primes for consecutive values of n, starting with n = 0.
import time
from utils import sieve_of_eratosthenes
if __name__ == '__main__':
t0 = time.time()
primes = set(sieve_of_eratosthenes(100000))
max_n = 0
result = (0, 0)
for a in xrange(-999, 1000):
if a % 200 == 0:
print '{}%'.format((100 * (a + 1000)) / 2000.0)
for b in xrange(-999, 1000):
n = 0
while True:
value = (n**2) + a*n + b
if value < 2 or value not in primes:
break
# n^2 + an + b, where |a| < 1000 and |b| < 1000
#
# where |n| is the modulus/absolute value of n
# e.g. |11| = 11 and |-4| = 4
# Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number
# of primes for consecutive values of n, starting with n = 0.
import time
from utils import sieve_of_eratosthenes
if __name__ == '__main__':
t0 = time.time()
primes = set(sieve_of_eratosthenes(100000))
max_n = 0
result = (0, 0)
for a in xrange(-999, 1000):
if a % 200 == 0:
print '{}%'.format((100 * (a + 1000)) / 2000.0)
for b in xrange(-999, 1000):
n = 0
while True:
value = (n**2) + a * n + b
if value < 2 or value not in primes:
break
#
# Solution to Project Euler Problem 50
#
import utils
prime_list = [x for x in utils.sieve_of_eratosthenes(1000000) if x < 100000]
max_ = 0
for i in range(0, len(prime_list)):
for j in range(i + 1, len(prime_list)):
if sum(prime_list[i:j]) > 1000000:
break
if utils.is_prime(sum(prime_list[i:j])) and j - i > max_:
max_ = j - i
print(str(sum(prime_list[i:j])) + ', ' + str(max_))
#
# Solution to Project Euler Problem 7
#
# What is the 10001st prime
#
import utils
# Brute force using the sieve_of_eratosthenes
if __name__ == "__main__":
primes = utils.sieve_of_eratosthenes(200000)
print(primes[10000])
def sum_of_primes():
return sum(utils.sieve_of_eratosthenes(2000000))
# The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
#
# Find the sum of all the primes below two million.
import time
from utils import sieve_of_eratosthenes
if __name__ == '__main__':
t0 = time.time()
results = sieve_of_eratosthenes(2000000)
runtime = time.time() - t0
print 'Runtime = {}'.format(runtime)
print 'Result = {}'.format(sum(results))
