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/

(\n->any ((==0).mod n) [3,5])

(\n->n `mod` 3 == 0 || n `mod` 5 == 0)

cleverly disguised with paranthesis