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 = [1] 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)

The Python file is available for download here.