 My solutions to the problems found at Project Euler.

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## Problem 21

```#! /usr/bin/python
# Problem: 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.
#
# Plan: Re-use old divsiors code from other problem, eliminate the number being divided from the list,
#       and the rest is just taking their definition of d, implementing it, and then counting.
#       Easy-peasy, nothing clever here.

import math

def divisors(n):
""" Return a list of divisors for n. """
sqrtn = int(math.sqrt(n))
divisors = []

if sqrtn ** 2 == n:
divisors.append(sqrtn)

for number in xrange(1,sqrtn):
if n % number == 0:
divisors.append(number)
divisors.append(n/number)

if n in divisors:
divisors.remove(n)

return divisors

def d(n):
return sum(divisors(n))

def amicable(a):
b = d(a)
if d(b) == a and not a == b:
return (a,b)

return False

if __name__ == "__main__":
numbers = []
for n in xrange(10000):
a = amicable(n)
if a:
if not a in numbers:
numbers.append(a)
if not a in numbers:
numbers.append(a)

print sum(numbers)

```