Project Euler Problem #1 - Multiples of 3 and 5 (in Scheme)

#lang scheme

; 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.

(define sum 0) ; sum of the multiples of 3 or 5 below 1000

; for i = 3 to 999
(define (findSumOfMultiples i)
  (cond 
    [(< i 1000) ; test all values less than 1000
      (cond
        [
          ; if i is a mod value of 0 in mod 3 or 5 (meaning i is a multiple of 3 or 5)
          (or (= (modulo i 3) 0) (= (modulo i 5) 0))
            (set! sum (+ sum i)) ; add i to the sum
        ]
      )
      (findSumOfMultiples (+ i 1)) ; call function again with incremented parameter value
    ]
    [else sum] ; hit 1000, so the sum of the multiples needs to be returned
  )
)

; call the function
(findSumOfMultiples 3)

And call it first with the parameter 3 because that's the least possible multiple of 3 or 5.

              Created: September 6, 2014
Completed in full by: Michael Yaworski