ProjectEuler21

2017/10/28

  • 問題

Problem 21:Amicable numbers
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.

  • 解答例
def sumDivisor(num):
    sum = 0
    i = 1
    while i * i < num:
        if num % i == 0:
            sum += i
            sum += (num // i)
        i += 1
    if i * i == num:
        sum += i
    return sum

sum = 0
for i in range(1, 10000):
    temp = sumDivisor(i)
    temp2 = sumDivisor(temp - i)
    if i == temp2 - (temp - i) and not(i == temp - i):
        sum += i

print(sum)