Solved Project Euler by clojure

Probrem1.clj

(apply + 
  (filter #(or (= (mod  % 3) 0) (=  (mod % 5) 0))
  (range 1 1000)))

Probrem2.clj

(defn fib [a b] (cons a (lazy-seq (fib b (+ b a)))))

(apply + 
  (filter even? 
    (take-while 
      #(< % 4000000) 
      (fib 1 2))))

Probrem3.clj

(defn primitive? [n]
  (if (< n 2)
    false
    (#(if (zero? (mod %1 %2))
            (= %1 %2)
            (recur %1 (inc %2))) n 2)))

((fn [i n] (if (<= i n)
  (let [mod-zero? (= (mod n i) 0) ret (/ n i)] 
    (if (and mod-zero? (is-primitive ret))
      (long ret)
      (if mod-zero?
        (recur 2 ret)
        (recur (inc i) n)))))) 2 600851475143)

Probrem4.clj

(defn palindrome? [n] (let [n-str (str "" n)] 
  (= n-str
    (clojure.string/join (reverse n-str)))))

(defn prod-3-digit? [n]
  (#(if (< %2 1000)
      (if (and (= 0  (mod %1 %2))
                  (< 99 (/ %1 %2)) (> 1000 (/ %1 %2)))
        true
        (recur %1 (inc %2)))
     false) n 100))


(first (filter #(and (palindrome? %) (prod-3-digit? %)) 
               (range (* 1000 1000) (* 99 99) -1)))

Probrem5.clj

(defn primitive? [n]
  (if (< n 2)
    false
    (#(if (zero? (mod %1 %2))
            (= %1 %2)
            (recur %1 (inc %2))) n 2)))

(defn get-max-pow [limit n] (#(if (> (* % %2) limit) %
                (recur (* % %2) %2)) n n))

((fn [n] (let [nums (filter primitive? (range 2 (inc n)))] 
                    (reduce 
          (fn [a b] (* a (get-max-pow n b)))
          1 nums))) 20)