Divisible by 3 or 5, in lisp

(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/

>>40
You forgot the paranthesis. LISP or GTFO.

divisibleBy3or5 = (\x -> (||) ((((.).(.)) (== 0) (mod)) (x) (3)) ((((.).(.)) (== 0) (mod)) (x) (5)))