def main(max_number):
summation = 0
prime_generator = generate_next_prime(start=5)
first_prime = prime_generator.next()
while first_prime <= max_number:
second_prime = prime_generator.next()
# we have the two following congruence relations:
# x = 0 (mod second_prime)
# x = first_prime(mod 10 ** (int(log10(first_prime)) + 1))
# so we use the Chinese Remainder Theorem to solve for x.
# thanks to Eigenray from http://projecteuler.net/thread=134
# for the suggestion to use CRT.
modulos = (second_prime, (10 ** (int(log10(first_prime)) + 1)),)
remainders = (0, first_prime,)
summation += crt(modulos, remainders)
#print first_prime, second_prime, residual
first_prime = second_prime
print "Sum(S) : %d" % summation
def main():
# it's hard to know whether a number even
# has a given number of prime factors
# (avg case is ln ln n). so, disallow
# the user from inputting a given number
# of prime factors to find like I might normally do.
repunit_length = 10 ** 9
number_of_primes_to_find = 40
prime_factors = set()
primes = generate_next_prime()
while len(prime_factors) != number_of_primes_to_find:
new_prime = primes.next()
# a base-10 repunit is formed via the following forumula:
# (10 ** (number of digits) - 1) / (10 - 1)
# we use modular exponentiation to see whether
# 10 ** (number of digits) is divisible by 9 * some prime
# with remainder 1.
if pow(10, repunit_length, 9 * new_prime) == 1:
prime_factors.add(new_prime)
print "Sum of first %d factors: %d" % (number_of_primes_to_find,
sum(prime_factors))
def main():
# it's hard to know whether a number even
# has a given number of prime factors
# (avg case is ln ln n). so, disallow
# the user from inputting a given number
# of prime factors to find like I might normally do.
repunit_length = 10**9
number_of_primes_to_find = 40
prime_factors = set()
primes = generate_next_prime()
while len(prime_factors) != number_of_primes_to_find:
new_prime = primes.next()
# a base-10 repunit is formed via the following forumula:
# (10 ** (number of digits) - 1) / (10 - 1)
# we use modular exponentiation to see whether
# 10 ** (number of digits) is divisible by 9 * some prime
# with remainder 1.
if pow(10, repunit_length, 9 * new_prime) == 1:
prime_factors.add(new_prime)
print "Sum of first %d factors: %d" % (number_of_primes_to_find,
sum(prime_factors))
def main(max_number):
summation = 0
prime_generator = generate_next_prime(start=5)
first_prime = prime_generator.next()
while first_prime <= max_number:
second_prime = prime_generator.next()
# we have the two following congruence relations:
# x = 0 (mod second_prime)
# x = first_prime(mod 10 ** (int(log10(first_prime)) + 1))
# so we use the Chinese Remainder Theorem to solve for x.
# thanks to Eigenray from http://projecteuler.net/thread=134
# for the suggestion to use CRT.
modulos = (
second_prime,
(10**(int(log10(first_prime)) + 1)),
)
remainders = (
0,
first_prime,
)
summation += crt(modulos, remainders)
#print first_prime, second_prime, residual
first_prime = second_prime
print "Sum(S) : %d" % summation
