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Divisible by 3 or 5, in lisp

Name: Anonymous 2007-08-12 17:20 ID:RyfBwU/E

(lambda (n) (or (= 0 (mod n 3)) (= 0 (mod n 5))))

how on earth do you write this in lisp, I mean better than I have here.. surely it must be possible?

Rube Goldberg devices are definitely acceptable if it makes the final code better.. hope you have some ideas /prog/

Name: Anonymous 2007-08-14 7:00 ID:iqrAANyA

divisibleBy3or5 x = divisibleBy x 3 || divisibleBy x 5
    where
    divisibleBy = (== 0) `dot` mod   
    dot = (.).(.)

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