Can you pack $53$ bricks of dimensions $1\times 1\times 4$ into a $6\times 6\times 6$ box?

Question

Can you pack 53 bricks of 1×1×4 size into 6×6×6 box?

Source: puzzledquant.com

My approach: I looked at the solution and visualized the box as $3d$ checkerboard. So in total we have $27$ such $2\times2\times2$ blocks. Now $14$ will be blue and $13$ red and each brick will be half blue and half red. So, total $52$ bricks will be used to fill it. Now, if it had been $6\times6\times7$ box, then too $52$ bricks only will be used. So, here we used the fact that $27$ is odd hence we divided into $14$ and $13$ and solved it. Is this correct? What if we had $28$ blocks then $14$ $14$ division and we would have fitted $14*4 = 56$ blocks here.

Answer 1

The original argument for the $6 \times 6 \times 6$ box is correct. Color the box as a $2 \times 2 \times 2$ checkerboard. There are $27\ 2 \times 2 \times 2$ cubes, $14$ black and $13$ white for a total of $112$ black cubes and $104$ white cubes. Each $1 \times 1 \times 4$ block takes two black and two white cubes, so $52$ blocks is the maximum that can be placed.

This argument does not suffice for a $6 \times 6 \times 7$ box because you have an additional layer which can accommodate another $8 \ 1 \times 1 \times 4$ blocks.

Answer 2

Yes.

Divide the 6 X 6 X 6 cube into 3 layers of 6 X 6 X 2. Fill up the middle layer such that there are twice as many 1 X 1 holes in it than in the outer 2 layers. Hint: look for symmetry here.

Only 2 of the holes in the outer 2 layers coincide and 4/6 of their lengths are filled up with a single brick so that the space left in each of them is 2/6 of the length (or half a brick size).

The rest of the cube is completely filled up.

Here is a screenshot of the solution: