Problem 20: Factorial digit sum

Problem:

Please find the problem here.

Solution:

It is well known that we can use Strling's approximation to approximate factorial, let see what we'll get from it. First of all, we have
$ \begin{eqnarray*} \sqrt{2 \pi} n^{n+\frac{1}{2}} e^{-n} \leq n! \leq e n^{n+\frac{1}{2}} e^{-n} \end{eqnarray*} $

Only the term $n^{n+\frac{1}{2}}e^{-n}$ matters, the rest are small. That gives us 100! is approximately $ 1.01 \times 10^{158} $.

This number is not small, but it is manageable to just compute 100! I don't have other solution to compute the sum of the digits anyway. Here is the code

namespace Euler
{
    using System;
    using System.Linq;
    using System.Numerics;

    internal static partial class Program
    {
        public static void Problem020()
        {
            BigInteger prod = new BigInteger(1);
            for (int i = 2; i <= 100; i++)
            {
                prod = BigInteger.Multiply(prod, new BigInteger(i));
            }
            Console.WriteLine(prod.ToString().Select(i => i - '0').Aggregate((x, y) => x + y));
        }
    }
}

Answer: 648

The value of 100! is actually
93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

It has 159 digits, so the Stirling approximation above is pretty close.