From the Project Euler

*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.*

**checked**

I found this the most difficult problem among the first 21, without using any theorem like that of Thabit ibn Qurra.

Here I found convenient to use the java class PriorityQueue.

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