;;;; ;;;;;;;;
;;;; Multiples of 3 and 5
;;;; 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.
;;;; ;;;;;;;;

(defvar *sum* 0)

(dotimes (x 1000)
  (when (or (eql 0 (mod x 5)) (eql 0 (mod x 3)))
    (incf *sum* x)))

(print *sum*)