# Find the sum of the digits in factorial of 100 in Erlang

This is a problem in poject Euler:

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!

I decided to solve it using erlang. This is what I've come up with.

-module(fact).
-compile(export_all).

fact(1) -> 1;
fact(N) -> N*fact(N-1).

tolist(N) when N < 10 -> [N];
tolist(N) when N >= 10 -> tolist(N div 10)++[N rem 10].

sum(List) -> sum(List, 0).
sum([H|T], C) -> sum(T, C + H);
sum([], C) -> C.


factfinds the factorial of the number. toList converts it to a list of integers and sum gets the sum of the integers in the list. I run this program like this:

fact:sum(fact:tolist(fact:fact(100))).


Do I have to use 3 functions? Is there a problem with me doing so?

Should I make another function that calls all 3 functions?

3. For extracting the numerical value from digit symbol - just subtract zero-value($0, number representation in Erlang). So, if accumulate all of this points, get: task_(N)-> lists:foldl(fun(X,Sum)->Sum+X-$0 end,0,integer_to_list(fact(N))).