sum.clj

; sums the number between 1 and 999 that are multiples of 3 or 5
(reduce + (filter #(= 0 (* (rem % 3) (rem % 5))) (take 999 (iterate inc 1))))

(or (= 0 (rem % 3)) (= 0 (rem % 5)))

can be reduced to

(= 0 (* (rem % 3) (rem % 5)))

but only because math is cool.

agreed, this is a good change and dries up the code. I was looking for a way to get rid of the duplicate declaration.

Of course that just works cause of a math trick. Now that I think about it, I think the more programming-languages way of drying it up would be
to replace

#(or (= 0 (rem % 3)) (= 0 (rem % 5)))

with

(fn [n](any? #%28= 0 %28rem n %%29%29 [3 5]))

This violates two of the rules you just sent me ;-)