Question

I am currently using the following code but its very slow for large numbers

        static int divisor(int number)
        {
            int i;
            for (i = number / 2; i >= 1; i--)
            {
                if (number % i == 0)
                {
                    break;
                }
            }
            return i;
        }

Answer 1

First thought you can find the smallest divisor d (not equal to 1 of course), then N/d will be the largest divisor you’re looking for.

For example if N is divisible by 3 then you’ll need 2 iterations to find the answer – in your case it would be about N/6 iterations.

Edit: To further improve your algorithm you can iterate through odd numbers only (after checking if you number is even) or, even better, if you have the list of primes pre-calculated then you can iterate through them only because smallest divisor is obviously is a prime number.