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/

(define (divisible n k) (= 0 (mod n k)))
(define (divisible-by-list n l) (fold (lambda (k o) (or o (divisible n k))) #f l))