## Euler Problem 20

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(); }

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Print | posted on Friday, June 17, 2011 5:22 PM |

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