Problem 20:

n! means n * (n – 1) * … * 3 * 2 * 1

For example, 10! = 10 * 9 * … * 3 * 2 * 1 = 3628800,
and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

Running time: 2ms

Assessment: Easy and fast using arbitrary-precision arithmetic.

import java.math.BigInteger;

public class Problem020
{
	private static int sumDigits(String s)
	{
		int sum = 0;
		
		for (int i = 0; i < s.length(); i++)
		{
			int j = Integer.parseInt(s.substring(i,i+1));
			sum += j;
		}
		
		return sum;
	}
	
	public static void main(String[] args)
	{
		long begin = System.currentTimeMillis();
		BigInteger fact = BigInteger.valueOf(1);
		
		for (int i = 1; i <= 100; i++)
			fact = fact.multiply(BigInteger.valueOf(i));
		
		long end = System.currentTimeMillis();
		
		System.out.println(sumDigits(fact.toString()));
		System.out.println(end-begin + "ms");
	}
}