欧拉问题21，为什么我的解决方案中会有多余的数字? [英] Euler problem 21, why are there extra numbers in my solution?

问题描述

以下是我的代码，用于解决 Euler#21，评估所有10000以下的友好数字之和.

The following is my code to solve Euler #21, Evaluate the sum of all the amicable numbers under 10000.

def divisor_sum(x):
    divisor_sum=0
    for i in range(1,x):
        if x%i==0:
            divisor_sum = divisor_sum + i
    return(divisor_sum)

for a in range (10000):
    if divisor_sum(divisor_sum(a)) == a:
        print(a)

响应为:

0
6
28
220
284
496
1184
1210
2620
2924
5020
5564
6232
6368
8128

我知道它想要总和，但是为了检查我的代码，我运行了要求和的数字，因为总和是错误的.

I know it wants the sums, but in order to check my code I ran what numbers it was summing since the sum was wrong.

奇怪的是，当我

The odd thing is that when I look online all of the numbers that should be here are found except 0, 6, 28, 496, 8128.

这是怎么回事?

推荐答案

好像您的程序正在找到完美数字以及友好数字:0、6、28、496和8128是前五个完美数字.请记住，友好数字是成对出现的.前五个完美数字没有，并且根据定义，不能有匹配的对，这会使它们成为友好的数字.

Looks like your program is finding perfect numbers as well as amicable numbers: 0, 6, 28, 496, and 8128 are the first five perfect numbers. Remember that amicable numbers come in pairs; the first five perfect numbers do not and by definition cannot have a matching pair that would make them an amicable number.

根据定义，一个完美数字的除数将求和回到完美数字，这意味着divisor_sum(divisor_sum(6)) == divisor_sum(6)

By definition, a perfect number's divisors will sum back to the perfect number, meaning divisor_sum(divisor_sum(6)) == divisor_sum(6)

正如评论者在我撰写本文时所指出的那样，您需要类似的内容:

As commenters pointed out as I was writing this, you need something like:

if divisor_sum(a)!=a

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