For problem 36, we need to find numbers that are palendromic in both base 2 and base 10. Thankfully, this is very simple to accomplish if we use the Java libraries to create the binary representations and also reuse a single function we wrote back in problem 4.
;; From Euler 4 (defn palindromic? [n] (= (seq (str n)) (reverse (str n)))) (defn binary-palindromic? [n] (palindromic? (Integer/toBinaryString n))) (defn euler-36  (reduce + (filter #(and (palindromic? %) (binary-palindromic? %)) (range 1 1000000))))
That was definitely too easy.