From: Todd Benson <caduceass@...>
Date: 2008-11-23T17:26:20+09:00
Subject: Re: Ayuda con Ejercicio

On Thu, Nov 20, 2008 at 11:21 AM,  <tirado.carlos@gmail.com> wrote:
> On Nov 19, 3:53 pm, Fernando Villafuerte <fernand...@hemac.com.mx>
> wrote:
>> lo que no se como hacer... es que el
>> resultado consiste solamente en la suma de los resultados
>> pero solo los que son números pares...
>
> Solo te hace falta agregar una variable para la suma, incrementandola
> cuando el numero es par (al dividirlo en 2 el residuo es cero):
>
>> a = 1
>> b = 2
>> stop = 4_000_000
>
> suma = 0
>
>> (1...500).each do
>>   s = a + b
>>   a = b
>>   b = s
>
> suma = suma + b if 0 == b.modulo(2)
>
>>   break if a > stop
>> end
>
> puts = suma

For fun (no guarantee of accuracy, but works for 4_000_000)...

include Math
R = Math.sqrt(5) #radical
P = (1 + R) / 2 #phi
my_number = 4_000_000
#inverting my_number to see approximately where n sits for F(n)
max_inverted = (log((my_number - 0.5) * R) / log(P)).to_i
#adding up F(n) for 0 to n
even_sum = (0..max_inverted).inject do |s, i|
  #next line is for direct F(n)
  #the inversion above is the opposite of this
  fib = (((P**i) - (1-P)**i) / R).floor
  #see if it's even
  fib % 2 == 0 ? (s + fib) : s
end
p even_sum


...It would be interesting to see what ceilings (like 4_000_000) this fails for.

cheers,
Todd