Cube1

[cinelli]

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Can a 6 x 6 x 6 cube be made with 27 bricks that are each 1x2x4 units? If so, how? If not, why not?

Cinelli

[Gengulphus]

Can a 6 x 6 x 6 cube be made with 27 bricks that are each 1x2x4 units? If so, how? If not, why not?

Spoiler...

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No, it cannot. To see why not, first split the cube into 27 2x2x2 subcubes in the obvious (and only) way, of two types A and B, with:

* The corner and face centre subcubes being of type A. There are therefore 8+6 = 14 subcubes of type A.

* The edge centre and body centre subcubes being of type B. There are therefore 12+1 = 13 subcubes of type B.

Or pictorially:

Top layer of subcubes
+-----+-----+-----+
|  A  |  B  |  A  |
+-----+-----+-----+
|  B  |  A  |  B  |
+-----+-----+-----+
|  A  |  B  |  A  |
+-----+-----+-----+

Middle layer of subcubes
+-----+-----+-----+
|  B  |  A  |  B  |
+-----+-----+-----+
|  A  |  B  |  A  |
+-----+-----+-----+
|  B  |  A  |  B  |
+-----+-----+-----+

Bottom layer of subcubes
+-----+-----+-----+
|  A  |  B  |  A  |
+-----+-----+-----+
|  B  |  A  |  B  |
+-----+-----+-----+
|  A  |  B  |  A  |
+-----+-----+-----+

Now colour the 8 1x1x1 cubelets making up each subcube of type A red and yellow as follows:

Top layer of cubelets
+-----+-----+
|  R  |  Y  |
+-----+-----+
|  Y  |  R  |
+-----+-----+

Bottom layer of cubelets
+-----+-----+
|  Y  |  R  |
+-----+-----+
|  R  |  Y  |
+-----+-----+

and the 8 1x1x1 cubelets making up each subcube of type A blue and green as follows:

Top layer of cubelets
+-----+-----+
|  B  |  G  |
+-----+-----+
|  G  |  B  |
+-----+-----+

Bottom layer of cubelets
+-----+-----+
|  G  |  B  |
+-----+-----+
|  B  |  G  |
+-----+-----+

The 216 cubelets making up the 6x6x6 cube are therefore coloured red, yellow, green and blue with 14*4 = 56 coloured each of red and yellow and 13*4 = 52 coloured each of green and blue.

However, it is easy to see that each layer of 6x6 cubelets making up the 6x6x6 cube, in each of the three directions (so 3*6 = 18 layers in total), is coloured:

+-----+-----+-----+-----+-----+-----+
|  W  |  X  |  Y  |  Z  |  W  |  X  |
+-----+-----+-----+-----+-----+-----+
|  X  |  W  |  Z  |  Y  |  X  |  W  |
+-----+-----+-----+-----+-----+-----+
|  Y  |  Z  |  W  |  X  |  Y  |  Z  |
+-----+-----+-----+-----+-----+-----+
|  Z  |  Y  |  X  |  W  |  Z  |  Y  |
+-----+-----+-----+-----+-----+-----+
|  W  |  X  |  Y  |  Z  |  W  |  X  |
+-----+-----+-----+-----+-----+-----+
|  X  |  W  |  Z  |  Y  |  X  |  W  |
+-----+-----+-----+-----+-----+-----+

for W, X, Y, Z some permutation of the four colours.

Now observe that however a 1x2x4 brick is placed, it is in one of those 18 layers, and however it is placed in such a layer, it contains 2 cubelets of each colour. So if I could construct the 6x6x6 cube from 27 such bricks, it would contain 27x2 = 54 cubelets of each colour - which it doesn't, so I cannot construct it that way.

Gengulphus

[cinelli]

Yes, It's an elegant solution. I think of this problem as being the 3D version of the familiar "mutilated chessboard" problem posed by Max Black in 1946.

Cinelli