Problem 21

问题描述：

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.

解决方法：

找到一个数(a)的因子的和(b)，然后这再找到这个和的因子的和(c)。

找到符合a==c的数。

 

 

public static int count_divisors_sum(long number) { //找到一个数的因子的 和
		int count = 1;
		for(int i=2; i*i<=number; i++){
			if(number%i==0){
				count+=i;
				if(i*i<number){
					count += number/i;
				}
			}
		}
		return count;
	}