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.