
# Problem 21:

# Let d(n) be defined as the sum of proper divisors of n

# (numbers less than n which divide evenly into n).

# If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair

# and each of a and b are called amicable numbers.

#

# For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20,

# 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284

# are 1, 2, 4, 71 and 142; so d(284) = 220.

#

# Evaluate the sum of all the amicable numbers under 10000.



# first find the proper divisors

import time


def find_divisors_sum(n):
    divisorList = []

    for i in range(1, n):

        if n % i == 0:
            divisorList.append(i)

    return sum(divisorList)


start = time.time()

amicable_sum = 0

for a in range(1, 10000):
    b = find_divisors_sum(a)
    if a == b:
        continue
    if find_divisors_sum(b) == a:
        amicable_sum += a

# if find_divisors_sum(i) == find_divisors_sum(find_divisors_sum(i)):

#        continue



elapsed = time.time() - start

print(amicable_sum)

print(elapsed)