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 l)
 (if (pair? l) #f
     (if (= (mod n (car l)) 0) #t
         (divisible n (cdr l)))))

(divisible n '(3 5))