**My solutions to the problems found at Project Euler. **

```
#! /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[0] in numbers:
numbers.append(a[0])
if not a[1] in numbers:
numbers.append(a[1])
print sum(numbers)
```