project euler: problem solutions

gistfile1.hrl

-module(euler).
-compile(export_all).

%% problem 1

%% If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
%% Find the sum of all the multiples of 3 or 5 below 1000.

%% Answer: 233168

sum_of_multiples(Sum, N, Max) ->
	Is_multiple = fun(X) -> (X rem 3) =:= 0 orelse (X rem 5) =:= 0 end,
	case N < Max of
		false -> Sum;
		true ->
			case Is_multiple(N) of
				true -> sum_of_multiples(Sum+N, N+1, Max);
				false -> sum_of_multiples(Sum, N+1, Max)
			end
	end.

problem_1() -> sum_of_multiples(0, 1, 1000).

%% problem 2

%% Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
%% 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
%% Find the sum of all the even-valued terms in the sequence which do not exceed four million.

%% Answer: 4613732

fib(1) -> 1;
fib(2) -> 2;
fib(N) when N > 2 -> fib(N-1) + fib(N-2).

is_even(N) when is_integer(N) -> (N rem 2) =:= 0.

sum_of_even_fib_terms(Sum, N, Max) ->
	Term = fib(N),
	case Term < Max of
		false -> Sum;
		true ->
			case is_even(Term) of
				false -> sum_of_even_fib_terms(Sum, N+1, Max);
				true -> sum_of_even_fib_terms(Sum+Term, N+1, Max)
			end
	end.

problem_2() -> sum_of_even_fib_terms(0, 1, 4000000).