A famous problem asks whether an 8x8 chessboard with two opposite corners deleted can be tiled with dominoes, where a domino is a rectangle congruent to two adjacent squares of the board.

Now, let C be an 8x8x8 cube with two diagonally opposite corners removed. For which integers $n>1$ is it possible to completely fill C using 1x1xn boxes in any orientation?

  • $\begingroup$ What a cool question! Thanks for posting. $\endgroup$ Commented Jun 16, 2022 at 20:39
  • $\begingroup$ Do you think it would be interesting to generalize your question by making C an mxmxm cube? $\endgroup$ Commented Jun 16, 2022 at 23:43

1 Answer 1


Our mutilated cube

has $8^3-2=510=2\cdot3\cdot5\cdot17$ little cubelets. This must be a multiple of $n$, and of course $n\leq8$, so for sure $n$ is one of {2,3,5,6}.

Let's first consider

$n=2$. Imagine a path on an ordinary 8x8 chessboard snaking from one corner to an adjacent corner. If we remove both of those corners then we can tile the resulting board with dominoes. So, begin on (let's say) the bottom face of our cube, and tile everything aside from the one missing cubelet (let's say at the SW corner) and the one in an adjacent corner (let's say the SE corner) with dominoes. Now add a vertical domino, and play the same game on the next layer, omitting the SE corner which we already covered and the NE corner. Now layer 3: NE/NW. Layer 4: NW/SW. 5: SW/SE. 6: SE/NE. 7: NE/NW. Now we come to the last layer, where we have already filled the NW corner and have removed the NE corner, and since these are adjacent everything is good. So $n=2$ is possible.

What about

$n=3$? Colour the cubelets in three colours according to the sum of their coordinates mod 3. If we put the removed corners on the $x=y=z$ diagonal then they are the same colour and we are left with 168 cubelets of that colour and 171 cubelets of each of the other two. But every 1x1x3 box contains one cubelet of each colour, so this is impossible. So $n=3$ is impossible, and so therefore is $n=6$.

Finally we must consider

$n=5$. Once again, colour cubelets according to $x+y+z$ mod 5, letting $(0,0,0)$ and $(7,7,7)$ be the cubelets removed. The resulting cubelet counts are: 100 of colour 0, 100 of colour 1, 103 of colour 2, 104 of colour 3, 105 of colour 4. And once again, every 1x1x5 box contains one of each colour, so this is impossible.

So the final answer is:

We can do it for $n=2$ (and, trivially, for $n=1$) but not for any other $n$.


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