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/


<?php
$x = 1000;
echo 1.5*(int)(($x-1)/3)*(int)(($x+2)/3) + 2.5*(int)(($x-1)/5)*(int)(($x+4)/5) - 7.5*(int)(($x-1)/15)*(int)(($x+14)/15);
?>
Everyone gtfo.