Project Euler, Problem 20: Factorial digit sum

$n! means n \times (n - 1) \times … \times 3 \times 2 \times 1$

For example, $10! = 10 \times 9 \times … \times 3 \times 2 \times 1 = 3628800$ ,
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 key to solving this one was to use strings to hold the products because they get big fast. To accomodate this I made two functions to help me multiply a string of numbers by another number.

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