Perfect squares and perfect cubes puzzle

Each of the 12 white squares is to be filled with a single digit. The four 2 digit numbers and the four 3 digit numbers thus created are either perfect squares or perfect cubes. These squares and cubes are all distinct and one of the down answers is both a square and a cube.

number-grid

Reveal Solution

We use $1d$ for “1 down”, etc.
The list of available squares and cubes is

$16, 25, 27, 36, 49, 64, 81, 100, 121, 125, 144, 169, 196, 216, 225, 256, 289,$

$324, 343, 361, 400, 441, 484, 512, 529, 576, 625, 676, 729, 784, 841, 900, 961.$
The two available numbers that are both square and cube are 64 and 729, and one of these is one of the down answers. Three possibilities lead to a contradiction, as we see next.
1. If then there are two cases according to .
  1. If $6a=625$ then $2d=512$ and $5d=256$ which implies that $7d=64=6d$ which won’t do.
  2. If $6a=676$ then there is no way to choose $2d$ .
2. If $2d=729$ then $6a=196$ and $5d=169$ and there is no way to choose $7a$ .
3. If $5d=729$ then, once again, there is no way to choose $7a$ .
  The only remaining possibility is $3d=64$ .
It follows that . But because can’t have 3 as a middle digit.
1. If $4a=484$ then $5d=841$ and $2d=144=6d$ , so this case fails.
2. If $4a=784$ then $5d=814$ and $6a$ ends in 64, are there are no 3 digit numbers in our list that fit.
Therefore $4a=144$ and $2d=512$ (since $2d=216$ leaves no possible $1a$ ). It follows that $1a=25$ . It quickly follows that $6a=324$ and $5d=441$ and we get to the completed grid as below.