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.
# 找到10000以下的所有亲密数,求他们的和。
def sum_factors(n): sum1 = 0 i = 1 temp = int(n**0.5+1) while i < temp: if n % i == 0: sum1 = sum1 + i + int(n/i) if n/i == i: sum1 -= i i += 1 return sum1 - n set1 = set() for a in range(1,10000): tempa = sum_factors(a) tempb = sum_factors(tempa) if tempb == a and a != tempa: set1.add(a) print(set1) print(sum(set1))结果: (220, 284) (1184, 1210) (2620, 2924) (5020, 5564) (6232, 6368)
sum: 31626