# Author: Deddryk
"""
Solution to problem 21.
"""
from utils.primes import sum_of_divisors
total = 0
for i in xrange(2, 10000):
sod = sum_of_divisors(i)
if sod > i and sum_of_divisors(sod) == i:
total += i + sod
print total
# Author: Deddryk
"""
Solution to problem 23.
This solution first generates a set of all abundant numbers.
It then loops through all numbers less than 28124, subtracting every abundant
number until it finds a result that is in the abundant set. If it does not
find a result, it is added to the running sum. Runs in 2.5 seconds on
my system. Could probably be improved
"""
from utils.primes import sum_of_divisors
abundants = set([x for x in xrange(1, 28124) if sum_of_divisors(x) > x])
non_abundant_sum = 0
for i in reversed(xrange(1, 28124)):
for num in abundants:
if i - num in abundants:
break
else:
non_abundant_sum += i
print non_abundant_sum
# Author: Deddryk
"""
Solution to problem 23.
This solution first generates a set of all abundant numbers.
It then loops through all numbers less than 28124, subtracting every abundant
number until it finds a result that is in the abundant set. If it does not
find a result, it is added to the running sum. Runs in 2.5 seconds on
my system. Could probably be improved
"""
from utils.primes import sum_of_divisors
abundants = set([x for x in xrange(1, 28124) if sum_of_divisors(x) > x])
non_abundant_sum = 0
for i in reversed(xrange(1, 28124)):
for num in abundants:
if i - num in abundants:
break
else:
non_abundant_sum += i
print non_abundant_sum
