PROJECT EULER #21

Link to 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.

using System;

namespace project_euler
{
class Program
{
    static void Main()
    {
        //Problem 21
        DateTime start = DateTime.Now;
        int sum=0;
        for (int i = 1; i < 10000; i++)
        {
            int n=IsAmicable(i);
            if (i!=n && i == IsAmicable(n))
                sum += i;
        }
        Console.WriteLine(sum);
        TimeSpan time = DateTime.Now - start;
        Console.WriteLine("This took {0}", time);
        Console.ReadKey();
    }
     public static int IsAmicable(int n)
    {
        int temp=0;
        for (int j = 1; j <= n/2; j++)
            if (n%j == 0)
                temp += j;
        return temp;
    }
}
}