Project Euler Problem 021

# Statement

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.

# Solution

Just a brute-force straight algorithm.

```def get_proper_divisors(n):
result = 
limit = int(n ** (0.5))
for i in range(2, limit):
if n % i == 0:
result.append(i)
result.append(n // i)
if n % limit == 0:
result.append(limit)
return result

def is_amicable(n):
pd = get_proper_divisors(n)
sum_pd = 0
for i in pd:
sum_pd += i
if sum_pd != n:
sum_pd2 = 0
for i in get_proper_divisors(sum_pd):
sum_pd2 += i
if sum_pd2 == n:
return True
return False

if __name__ == '__main__':
result = 0
for i in range(1, 10000):
if is_amicable(i):
result += i
print("The result is:", result)
```