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racket solutions for 1 and 2

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Lowl3v3l 2017-05-09 19:20:46 +02:00
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commit faf8f4962a
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#lang racket
(define (multiples limit)
(foldl + 0 (filter (lambda (x) (cond
[(= 0 (remainder x 3)) #t]
[(= 0 (remainder x 5)) #t]
(else #f)))
(range limit))))
(multiples 1000)