I'm learning Clojure and solved Project Euler #1, I'd like to have my code reviewed on all aspects and have one specific question. This is also my first Functional Programming language, I have however used some form of FP in Java and Python before.

Task: Multiples of 3 and 5

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

My solution

 (ns euler)

 (defn is-divisible-by?
   [x divisor]
   (zero? (mod x divisor)))

(->> (range 1000)
     (filter #(or (is-divisible-by? % 3) (is-divisible-by? % 5)))
     (reduce +)

My additional question is whether the filter part can be written more succintly as I'm duplicating code right now while all I need to actually to is dos ome operation on 3 and 5.


1 Answer 1


Good to extract the method is-divisible-by?. Ending with a question mark indicates a predicate, so is- prefix is not necessary. I think it is somewhat more readable if the divisor comes before the value, but that might be personal preference.

(defn divisible-by?
  [divisor x]
  (zero? (mod x divisor)))

Furthermore, if you provide a conditional in the fn to reduce you don't need a filter. It is debatable if it is more readable, but it is shorter and saves one walk over the collection I think (although I measure no difference in performance between the if-statement and the filter). Also there is no need for parentheses around println:

(->> (range 1000)
     (reduce #(if (or (divisible-by? 5 %2)
                      (divisible-by? 3 %2))
                (+ %1 %2)

You can get inspired by other answers on the projecteuler thread for question 1.


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