$6bcde$ is a five-digit number. If $6-d=x, c=y$ and $xy$ is a two digit square number in which each digit is a square number. If $b = p$, $e = q$ and $pq$ is a two-digit perfect square number in which each digit cube, the cube root of $6bcde$ is?
This question was on an exam I had taken about a month ago. Even now, I can't find the answer to this.
If $xy$ is a two-digit square number in which each number is a square, then $xy = 49$; this is the only such number. So $x = 4$ and $y = 9$. Similarly, the only two-digit square number in which each digit is a cube is 81, so $p = 8$ and $q = 1$. All together, then, the number $6bcde$ is 68921, and $\sqrt[3]{68921} = 41.$