Created
June 28, 2010 22:46
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| ; 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)))) |
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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 ;-)