def main():
LIMIT = 10000
divisors_sum = {number: sum(integers.proper_divisors(number))
for number in range(2, LIMIT + 1)}
amicable_numbers = []
for n1, n2 in itertools.product(divisors_sum, divisors_sum):
# if the sum of proper divisors of n_1 is the same as n_2 and vice
# versa, then n_1 is an amicable number.
if divisors_sum[n1] == n2 and divisors_sum[n2] == n1 and n1 != n2:
amicable_numbers.append(n1)
answer = sum(amicable_numbers)
return answer
def main():
# Lowest abundant number is 12
# Every number greater than 28123 can be written as the sum of two abundant
# numbers so only need to check up to 28123.
LOWERBOUND = 12
UPPERBOUND = 28123
abundant_numbers = [number for number in range(LOWERBOUND, UPPERBOUND + 1)
if sum(integers.proper_divisors(number)) > number]
# Get the cross product of abundant numbers so we can determine every
# number less than 28123 that is a sum of two abundant numbers.
cross_product = itertools.product(abundant_numbers, abundant_numbers)
sum_of_abundants = set((sum(tup) for tup in cross_product if
sum(tup) <= UPPERBOUND))
# Every number less than or equal to UPPERBOUND that is not a part of the
# set sum_of_abundants is the set we need.
not_sum_of_abundants = set(range(1, UPPERBOUND + 1)) - sum_of_abundants
answer = sum(not_sum_of_abundants)
return answer