FizzBuzz solved using condp function

fizzbuzz.clj

(defn FizzBuzz
  "Returns 'Fizz', 'Buzz' or 'FizzBuzz' if n is divisible by
  3, 5 or both, respectively, or else just n."
  [n]
  (condp #(zero? (mod %2 %1)) n
    15 "FizzBuzz"
     3 "Fizz"
     5 "Buzz"
       (str n)))

(->>
  (range 1 16)
  (map FizzBuzz)
  (println))