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/

>>59

divisiblebyany l x  = product (map (mod x) l) == 0

I still think that

divisiblebyany l x = any ((== 0).(mod x)) l

or
divisiblebyany x = any ((== 0).(mod x))

or

divisiblebyany x = any (divisibleby x)

is better though.