The next problem up in Project Euler concerns a rather exotic-sounding kind of number pairing: the amicable pair. The problem itself contains a thorough definition and explanation of what defines such a pair:

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.

To find all amicable numbers under 10,000, I used the algorithm from my solution to problem 12 to find the divisors for all numbers between 1 and 9,999, and then searched an array constructed therefrom for amicable pairs.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace Problem021
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] pairs = new int[10000];
            int divisors = 1;
            int sumOfAmicableNumbers = 0;

            for (int i = 1; i < 10000; i++)
            {
                for (int j = 2; j <= (int)Math.Sqrt(i); j++)
                {
                    if (j == Math.Sqrt(i))
                        break;
                    if (i % j == 0)
                        divisors += j + (i / j);
                }
                pairs[i] = divisors;
                divisors = 1;
            }

            for (int i = 1; i < 10000; i++)
            {
                if (pairs[i] < 10000)
                {
                    if (i != pairs[i] && i == pairs[pairs[i]])
                    {
                        sumOfAmicableNumbers += i;
                    }
                }
            }

            Console.WriteLine(sumOfAmicableNumbers);
        }
    }
}