def problem_50():
limit = 10**6
# find the smallest list of primes which will sum up to at least the limit we set above
for x in count(1):
primes = list(get_primes_under(10**x))
if sum(primes) > limit:
break
# starting with the longest possible sequence (every element in the list), and then working
# to smaller sequence lengths
for length in reversed(range(1, len(primes) + 1)):
# Use a "sliding window" approach. Starting at the low end of the list, grab the first
# `length` elements and check it. Then slide the window right 1 element, and grab those,
# and check it, etc
for start in range(len(primes)-length):
candidate = sum(primes[start:start+length])
# if the candidate sum is bigger than the limit, no point in checking any more windows
# of size `length`. Break the inner loop, and reduce the length
if candidate >= limit:
break
# if the candidate sum is prime, we're done!
if is_prime(candidate):
print(candidate)
return
def problem_49():
# get a generator of primes under 10k (since we know the primes in question are 4 digits)
# and then discard all 3-digit ones
primes = get_primes_under(10**4)
while next(primes) < 999:
pass
# turn the remainder of the primes in the generator into a list
primes = list(primes)
for p in primes:
if p in (1487, 4817, 8147):
# skip this one since the problem is asking for the other one
continue
# get the digits of the current prime in sorted order, and get any other primes
# in the list which contain exactly the same digits
sorted_digits = sorted(str(p))
others = [x for x in primes if x != p and sorted(str(x)) == sorted_digits]
# if we don't have at least 2 other primes with the same digits, skip to the next prime
if len(others) < 2:
continue
# check each pair of primes in the `others` list, and if the difference between them is
# the same as the difference between our current prime and the first of these, we have
# a winner
for n1 in others:
for n2 in [x for x in others if x > n1]:
if (n2 - n1) == (n1 - p):
print('{}{}{}'.format(p, n1, n2))
return
def can_be_written_as_prime_plus_twice_square(n):
""" Returns True if n can be written as the sum of a prime and twice a square number, or
returns False otherwise. """
if is_prime(n):
return True
# For every prime under n, find what value of x satisfies: n = prime + 2x^2
# If any of those values x are an integer, then n can be written as prime + 2x square number
for prime in get_primes_under(n):
x = ((n - prime) / 2) ** 0.5
if x % 1 == 0:
return True
# If we get here, n cannot be written as prime + 2x square number
return False
def problem_60():
# get list of primes, discard 2
primes = list(get_primes_under(9000))
primes.pop(0)
for a in primes:
# build up a list of primes which pair with `a` to concatenate and form other primes
pairs_with_a = list()
for b in primes:
if a == b: continue
if concats_to_prime(a, b):
pairs_with_a.append(b)
# For every combination of 4 in the primes which pair with `a`:
# if every pair of them also concats_to_prime, we found the winner!
#
# We know the first result obtained must be the correct answer, since combinations()
# yield combos in lexographical order, so we'll get combos with smaller numbers first
for combo in combinations(pairs_with_a, 4):
if all(concats_to_prime(p1, p2) for p1, p2 in product(combo, repeat=2) if p1 != p2):
print(sum(combo) + a)
return
def problem_10():
print(sum(get_primes_under(2000000)))
644 = 2² × 7 × 23
645 = 3 × 5 × 43
646 = 2 × 17 × 19.
Find the first four consecutive integers to have four distinct prime factors. What is the first
of these numbers?
"""
from utils.timer import time_it
from utils.primes import get_primes_under
from itertools import count
#-------------------------------------------------------------------------------------------------
# this limit determined experimentally
primes = list(get_primes_under(10**3))
def has_4_distinct_prime_factors(n):
""" Returns True if n has exactly 4 distinct prime factors, or False otherwise. """
factor_count = 0
for p in primes:
if n % p == 0:
# if it already has 4 factors so far, another will disqualify it, so just return False
if factor_count == 4:
return False
factor_count += 1
# we don't have any more than 4 here, but make sure we don't have less
return factor_count == 4
