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.

+-;nats
(defn nats [] 
  (iterate inc 1)
)

(defn p1 []
  (println
    (apply + (filter #(or (= 0 (rem % 3)) (= 0 (rem % 5))) (take 999 (nats))))
  )
)

p1()->
	p1(1,0).
p1(1000,Sum)->io:format("~w~n",[Sum]);
p1(C,Sum)->
	if
		C rem 5 == 0; C rem 3 == 0->
			p1(C+1, Sum+C);
		true->
			p1(C+1, Sum)
	end.

+-#Enumerable
module Enumerable
  def sum
    self.inject{|u,v|u+v}
  end
  def product
    self.inject{|u,v|u*v}
  end
  def count(ob)
    self.select{|i|i==ob}.length
  end
  def take_while
    return [] unless yield(self.first)
    if(self.class==Range)
      evalPoint=self.first
      evalPoint=evalPoint.succ while yield(evalPoint.succ) and not evalPoint==self.last
      return self.first..evalPoint
    else
      s=self.to_a
      upTo=1
      upTo+=1 while yield(s[upTo]) and upTo<self.length
      return self[0...upTo]
    end
  end
end

def p1
  puts (1...1000).select{|i|i%3==0 or i%5==0}.sum
end

def p1 {
  var x=0
  for(i <- (1 until 1000) if i%3==0 || i%5==0) x+=i
  println(x)
}