This was probably one of the easiest ones to complete – a quick bash got me the following…
The Problem 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!
The Solution
private static BigInteger Factorial(int num)
{
if (num > 1) return (BigInteger)num * Factorial(num - 1);
else return 1;
}
private static BigInteger SumDigits(string digits)
{
BigInteger result = 0;
foreach (char number in digits)
{
result += Convert.ToInt32(number)-48;
}
return result;
}
static void Main(string[] args)
{
Console.WriteLine(SumDigits(Factorial(100).ToString()));
Console.ReadLine();
}
