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/

(defun % (d)
 (lambda (n) (= 0 (mod n d))))

(defun all (&rest predicates)
 (lambda (x)
  (every (function identity)
   (mapcar (function (lambda (p) (funcall p x))) predicates))))

(all (% 3) (% 5))