04 Jul '06 10:34

Is it possible to fill a 6x6x6 cube with 1x1x4 cuboids? (No overlaps or gaps, arranged vertically and horizontally.)

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04 Jul '06 16:47

No.*Originally posted by SPMars***Is it possible to fill a 6x6x6 cube with 1x1x4 cuboids? (No overlaps or gaps, arranged vertically and horizontally.)**

Imagine the 6x6x6 cube as composed of 27 2x2x2 cubes coloured in chess board fashion - ie no cubes of the same colour adjacent. There would be 5+4+5 (=14) of one colour and 4+5+4 (=13) of the other.

However, no matter how you place the 1x1x4 pieces they have to take up and equal amount of each colour. Clearly a contradiction.- Joined
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Isle of Misfit Toys05 Jul '06 14:46

Wait... no. Why imagine the cube as composed of 27 2x2x2 pieces? Especially when the given piece is 1x1x4? A 2x2x2 piece is 8 little (1x1x1) cubes, while the given 1x1x4 piece is 4 little cubes. By halving the number of pieces, you have a 50-50 chance of creating an odd number of pieces.*Originally posted by howardbradley***No.**

Imagine the 6x6x6 cube as composed of 27 2x2x2 cubes coloured in chess board fashion - ie no cubes of the same colour adjacent. There would be 5+4+5 (=14) of one colour and 4+5+4 (=13) of the other.

However, no matter how you place the 1x1x4 pieces they have to take up and equal amount of each colour. Clearly a contradiction.

Clearly, you can't fit 54 1x1x4 pieces into a cube 6x6x6.

But the logic you describe has nothing to do with it.

Using similar logic, I just as easily could have said: Imagine the 6x6x6 cube as composed of 8 3x3x3 cubes colored in chess board fashion - ie no cubes of the same color adjacent. There would be 2+2 (=4) of one color and 2+2 (=4) of the other. No matter how you place the 1x1x4 pieces they have to take up an equal amount of each color. 4=4, so therefore, it is possible.

Clearly, this isn't the case.- Joined
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05 Jul '06 19:462 edits

Hmmmmmm.*Originally posted by Suzianne***Wait... no. Why imagine the cube as composed of 27 2x2x2 pieces? Especially when the given piece is 1x1x4? A 2x2x2 piece is 8 little (1x1x1) cubes, while the given 1x1x4 piece is 4 little cubes. By halving the number of pieces, you have a 50-50 chance of creating an odd number of pieces.**

Clearly, you can't fit 54 1x1x4 pieces into a cube 6x6x6.

Bu ...[text shortened]... ual amount of each color. 4=4, so therefore, it is possible.

Clearly, this isn't the case.

Perhaps in the land of "Know-it-all" statements like this: "Clearly, you can't fit 54 1x1x4 pieces into a cube 6x6x6." constitute proof. Not here though.

I didn't have a 50-50 chance of creating an odd number of pieces. I had a 100% chance; because I chose that subdivision*precisely because it did*create an odd number of pieces. And thereby more of one colour than another.

It's a technique called*Reductio ad absurdum.*

Those of you who don't already know everything may be interested in this article: http://en.wikipedia.org/wiki/Reductio_ad_absurdum. It has a very apposite quote from G H Hardy at the end.

PS. This statement "No matter how you place the 1x1x4 pieces they have to take up an equal amount of each color." is wrong for your 8 3x3x3 cubes. Can you see why?

That's right: you could put a 1x1x4 so that 3 of its cubes were one colour and 1 cube the other. 3 does not equal 1 (even for large values of 1.)- Joined
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05 Jul '06 21:322 editsNeedless to say howardbradley, your argument was logically sound.

Suzianne -- not sure where you were coming from with your post...sorry!

The problem is very similar to "not being able to tile a chessboard with 2x1 rectangles in such a way that one pair of opposite corners is omitted". This is because opposite corners always have the same colour, whereas a 2x1 tile necessarily covers 1 white square and 1 black square.- Joined
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p^2.sin(phi)