Packing 53 Bricks of Size 1x1x4 Into a 6x6x6 Box
You have a cubic box with internal dimensions $6 \times 6 \times 6$ (volume $216$) and 53 rigid bricks, each of dimensions $1 \times 1 \times 4$ (total volume $53 \times 4 = 212$). Bricks must be placed with their edges parallel to the box's edges, so each brick occupies four unit cubes in a straight line along one of the three axes.
Can all 53 bricks be packed inside the box? Prove your answer.
Hints
- Volume alone leaves 4 unit cubes of slack, so a volume argument cannot settle it. Think about the classic mutilated chessboard: colour the cells so that every brick covers a fixed pattern of colours.
- Do not colour unit cubes individually; colour the 27 blocks of size $2 imes 2 imes 2$ that tile the box in a 3D checkerboard pattern. How many blocks of each colour are there, and how many unit cubes of each colour does one brick cover?
- There are 14 blocks of one colour and 13 of the other, so 104 unit cubes of the minority colour. Every $1 imes 1 imes 4$ brick covers exactly 2 unit cubes of each colour, so at most $104/2 = 52$ bricks fit.
Worked Solution
How to Think About It: A packing question with a tiny volume slack is a signal to look for a colouring invariant, as in the mutilated chessboard. We need a colouring such that (i) every brick covers the same number of cells of each colour, and (ii) the box has an unequal number of cells of each colour. The mismatch then bounds how many bricks can fit.
Approach: Tile the box with 27 blocks of size $2 \times 2 \times 2$, checkerboard-colour those blocks, count cells per colour, and count how each brick splits across colours.
Formal Solution:
*Step 1 -- The colouring.* Index unit cubes by $(x, y, z)$ with $x, y, z \in \{0, 1, \ldots, 5\}$. Partition the box into $27$ blocks of size $2 \times 2 \times 2$; the block containing $(x, y, z)$ is $(\lfloor x/2 \rfloor, \lfloor y/2 \rfloor, \lfloor z/2 \rfloor) \in \{0,1,2\}^3$. Colour a block black if $\lfloor x/2 \rfloor + \lfloor y/2 \rfloor + \lfloor z/2 \rfloor$ is odd and white if it is even, and give every unit cube the colour of its block.
*Step 2 -- Count cells per colour.* Among the 27 blocks, the parity of the coordinate sum is even for 14 blocks and odd for 13 blocks (the corner blocks such as $(0,0,0)$ and $(2,2,2)$ are even, and a $3 \times 3 \times 3$ checkerboard has one more of the corner colour). So there are $14 \times 8 = 112$ white cells and $13 \times 8 = 104$ black cells.
*Step 3 -- What one brick covers.* A brick occupies 4 consecutive cells along one axis, say $(x, y, z), (x+1, y, z), (x+2, y, z), (x+3, y, z)$ with $x \in \{0, 1, 2\}$. As $x$ runs over four consecutive values, $\lfloor x/2 \rfloor$ takes two consecutive values, each exactly twice: for $x_0 = 0$ we get $0, 0, 1, 1$; for $x_0 = 1$ we get $0, 1, 1, 2$; for $x_0 = 2$ we get $1, 1, 2, 2$. Adjacent blocks along an axis have opposite colours, so every brick covers exactly 2 black cells and 2 white cells. (A direct check over all $324$ possible brick placements confirms this.)
*Step 4 -- The bound.* If $k$ bricks are packed, they cover $2k$ black cells, and there are only $104$ black cells. Hence $2k \le 104$, so $k \le 52$.
*Step 5 -- Conclusion.* 53 bricks would need 106 black cells, but the box has only 104. No packing exists. (The volume slack of 4 cells was a red herring: what actually binds is the colour imbalance, and 52 bricks is the true ceiling.)
Answer: No. Checkerboard-colouring the 27 sub-blocks of size $2 \times 2 \times 2$ gives 104 cells of one colour and 112 of the other; every $1 \times 1 \times 4$ brick covers exactly 2 cells of each colour, so at most $104 / 2 = 52$ bricks fit, and 53 cannot.
Intuition
Volume says there is room, but an invariant says there is not: any straight run of 4 unit cubes crosses two adjacent $2 \times 2 \times 2$ blocks and therefore uses 2 cells of each checkerboard colour, while the box itself has an unequal split of colours (104 vs 112). The scarce colour caps the count at 52. Colouring and parity invariants are the standard way to prove impossibility in tiling and packing problems, and the same mindset (find a conserved quantity that the desired outcome would violate) is what you use to show that a proposed trade or state cannot be reached from where you are.