Box & Bricks

There is a colouring proof to show this is impossible.

Consider the $10 \times 10 \times 10$ cube broken down into 125 $2 \times 2 \times 2$ cubes. Colour these in a black and white alternating fashion such that no small cubes which are adjacent to each other are the same colour. It should look something like this (except with more cubes per side):

Assume that the corners are all white. Then, there are 63 white $2 \times 2 \times 2$ cubes and 62 black $2 \times 2 \times 2$ cubes.

Note that each $1 \times 2 \times 4$ blocks will have exactly 4 unit cubes white and 4 unit cubes black. Thus, if 125 bricks are used, the number of white and number of black unit cubes must be the same, which is impossible as there is 1 more white $2 \times 2 \times 2$ cube than black.

Therefore, it is impossible.

According to De Brujin's Theorem , a harmonic brick of dimensions $a\times ab\times abc$ can be packed in a box if & only if a box has dimensions of $ap\times abq\times abcr$ for some integers $a, b, c, p, q, r$ . In other words, for every side length of the box, it must be a multiple of one of the brick's side length exclusively.

However, in the question, since the cube's length $10$ is not a multiple of $4$ , the bricks can not be perfectly packed into this cubic box.

By using proportion, we have:

$\left\lfloor \cfrac{10}{1} \right\rfloor \times \left\lfloor \cfrac{10}{2} \right\rfloor \times \left\lfloor \cfrac{10}{4} \right\rfloor \ =?= \ 125 \\ 10 \times 5 \times 2 \ =?= \ 125 \\ 100 \neq 125$

Hence, the 125 pink bricks can't be perfectly placed into a box of dimensions 10 * 10 * 10.