You are given 27 pieces of 1x2x4 cuboids. Is it possible to build a 6x6x6 cube using those 27 cuboids?

  • $\begingroup$ Do you want theoretical or empirical answers? I think theoretical would be better. $\endgroup$
    – smci
    Dec 31, 2019 at 12:45

1 Answer 1


If I'm not mistaken, I believe that it is



enter image description here

Using this diagram, each cuboid will cover exactly one RED X.
As there are $28$ RED X, thus we need at least $28$ cuboids which is not possible.

  • $\begingroup$ Surely, this is a very nice and unique tiling puzzle nevertheless! :D $\endgroup$
    – athin
    Dec 31, 2019 at 0:02
  • $\begingroup$ I believe this doesn't prove that it is impossible. There could be a placement with 27 X. $\endgroup$ Dec 31, 2019 at 2:38
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    $\begingroup$ @DmitryKamenetsky After placing 26 cuboids, you will have filled exactly 26 of the Xs. The space you have left cannot be filled with the 27th cuboid since it contains two of the Xs. $\endgroup$ Dec 31, 2019 at 9:39
  • 3
    $\begingroup$ @LorenPechtel The images are not faces, they are layers, stacked on on the other. If you look closely you will see that the cuboids can't double up on x's even between different layers. $\endgroup$
    – TyJ
    Dec 31, 2019 at 20:35
  • 2
    $\begingroup$ Here is an equivalent but slightly different colouring argument. Think of the 6x6x6 cube as being built from 27 2x2x2 blocks. Give this a checkerboard colouring, so that there are 14 white blocks and 13 black ones. Every cuboid piece, however it is placed, will consume an equal amount of the two colours. You will therefore always have 8 unfilled white spaces. $\endgroup$ Jan 1, 2020 at 7:05

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