def get_prime(n):
return primesieve.get_primes(n)[-1]
'''
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
'''
from primesieve import get_primes
if __name__ == '__main__':
num = 600851475143
# Largest distinct factor is <= to SQRT
max = round(pow(600851475143, .5))
primes = get_primes(max)
# Since we're looking for the largest, start at the end
primes.reverse()
for p in primes:
if num % p == 0:
print p
break
"""
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 primesieve import get_primes
if __name__ == "__main__":
index = 10001
primes = get_primes(1000000)
print primes[index - 1]
'''
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
'''
from primesieve import get_primes
import math
if __name__ == '__main__':
max = 20
numbers = range(2,max+1,1)
primes = get_primes(numbers[-1])
result = 1
# Per the fundamental theorem of arithmetic:
# http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
# Every number can be factored as a multiple of primes in the form:
# p1 ^ i1 * p2 ^ i2 * p3 ^ i3 ... pi ^ in
# Therefore, the only numbers we need to multiply are the primes, but
# the tricky part is to figure out the specific exponents of each prime
# To do that we would need to determine the largest exponent of our prime
# that would "fit" in the number
for p in primes:
# We need to find the smallest integer that satisfies this
# p^x <= max <= p^(x+1)
# x * log(p) = log(max)
# x = log(max) / log(p)
# x = floor(log(max)/log(p)
def test_primes(self):
self.assertEquals(get_primes(0), [])
self.assertEquals(get_primes(1), [])
self.assertEquals(get_primes(2), [2])
self.assertEquals(get_primes(3), [2,3])
self.assertEquals(get_primes(20), [2,3,5,7,11,13,17,19])
def __init__(self, max):
# Store all the primes up to the max number to speed up calls
for p in get_primes(max):
self.primes[p] = True
'''
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
'''
from primesieve import get_primes
if __name__ == '__main__':
print sum(get_primes(2000000))
def get_prime(n):
return primesieve.get_primes(n)[-1]
