Best way to find all factors of a given number

The % (remainder) operator is the one to use here. If x % y == 0 then x is divisible by y. (Assuming 0 < y <= x)

I'd personally implement this as a method returning an IEnumerable<int> using an iterator block.

pseudocode:

• Loop from 1 to the square root of the number, call the index "i".
• if number mod i is 0, add i and number / i to the list of factors.

realocode:

public List<int> Factor(int number) 
{
    var factors = new List<int>();
    int max = (int)Math.Sqrt(number);  // Round down

    for (int factor = 1; factor <= max; ++factor) // Test from 1 to the square root, or the int below it, inclusive.
    {  
        if (number % factor == 0) 
        {
            factors.Add(factor);
            if (factor != number/factor) // Don't add the square root twice!  Thanks Jon
                factors.Add(number/factor);
        }
    }
    return factors;
}

As Jon Skeet mentioned, you could implement this as an IEnumerable<int> as well - use yield instead of adding to a list. The advantage with List<int> is that it could be sorted before return if required. Then again, you could get a sorted enumerator with a hybrid approach, yielding the first factor and storing the second one in each iteration of the loop, then yielding each value that was stored in reverse order.

You will also want to do something to handle the case where a negative number passed into the function.

Very late but the accepted answer (a while back) didn't not give the correct results.

Thanks to Merlyn, I got now got the reason for the square as a 'max' below the corrected sample. althought the answer from Echostorm seems more complete.

public static IEnumerable<uint> GetFactors(uint x)
{
    for (uint i = 1; i * i <= x; i++)
    {
        if (x % i == 0)
        {
            yield return i;
            if (i != x / i)
                yield return x / i;
        }
    }
}