A simple solution to FizzBuzz — Clojure

fizzbuzz.clj

(defn fizzbuzz?
	"Determines what to print for a given number"
	[x]
	(condp #(zero? (mod %2 %1)) x
		15 "fizzbuzz"
		3 "fizz"
		5 "buzz"
		1 x))

(defn fizzbuzz
	"Prints fizzes and buzzes for numbers between 1 and x"
	[x]
	(doseq [i (range 1 (inc x))]
		(println (fizzbuzz? i))))

Notes:

(rem) can be used in place of (mod) with no change to the code — the difference in efficiency seems negligible.

(and (mod 3 x) (mod 5 x)) == (mod 15 x) because a number is divisible by 3 and 5 ⟷ the number is divisible by 15.

I prefer doseq, but dotimes can be used instead in the body of fizzbuzz:

(dotimes [i x]
    (println (fizzbuzz? (inc i))))