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;;;; ;;;;;;;;
;;;; 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*)
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