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Topic: A perfect cube built with bricks? (Read 1106 times) |
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ecoist
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A perfect cube built with bricks?
« on: Apr 2nd, 2006, 10:45am » |
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Can a perfect 12-inch cube be built with 27 2- by 4- by 8-inch bricks?
Sorry, guys. I'm unable to recall my original solution.
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Grimbal
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Re: A perfect cube built with bricks?
« Reply #1 on: Apr 2nd, 2006, 2:53pm » |
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I'd say it cannot be done.
Consider a cube 6x6x6 to be filled by 1x2x4 pieces.
Label the cubes in the 6x6x6 cube as follows:
ABCDAB BCDABC CDABCD DABCDA ABCDAB BCDABC
BCDABC CDABCD DABCDA ABCDAB BCDABC CDABCD
CDABCD DABCDA ABCDAB BCDABC CDABCD DABCDA
DABCDA ABCDAB BCDABC CDABCD DABCDA ABCDAB
ABCDAB BCDABC CDABCD DABCDA ABCDAB BCDABC
BCDABC CDABCD DABCDA ABCDAB BCDABC CDABCD
You can see that every block covers 2 of each type of cube.
But there are only 53 A's anc D's, and 55 B's and C's.
More simply, a block always covers 2 A's, but there is an odd number of them.
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ecoist
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Re: A perfect cube built with bricks?
« Reply #2 on: Apr 3rd, 2006, 6:15am » |
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Let's see if I understand Grimbal's solution. His argument also seems to show that 1 by 4 polyominoes cannot exactly cover a 6x6 checkerboard, that a 6x6x6 cube cannot be exactly filled by 1x1x4 bricks, and that, maybe, a 6x6x6x6 hypercube can be exactly filled by 1x1x4 bricks.
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Grimbal
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Re: A perfect cube built with bricks?
« Reply #3 on: Apr 3rd, 2006, 6:57am » |
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Exactly, except of that 1x1x4 blocks won't fill any hypervolume... 
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Barukh
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Re: A perfect cube built with bricks?
« Reply #4 on: Apr 3rd, 2006, 8:47am » |
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The following thread discusses (IMHO cleverly) a similar problem.
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ecoist
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Re: A perfect cube built with bricks?
« Reply #5 on: Apr 3rd, 2006, 9:51am » |
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Ha, ha! How about a 1x1x1x4 brick? Thanks, Barukh. Got a little carried away. Grimbal's solution is a gem!
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Barukh
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Re: A perfect cube built with bricks?
« Reply #6 on: Apr 3rd, 2006, 11:53pm » |
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on Apr 3rd, 2006, 9:51am, ecoist wrote:| How about a 1x1x1x4 brick? |
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The same mathematician (G. de Bruijn) proved that a hyper-box can be filled with hyper-bricks 1x1x...x1xn if and only if at least one of the dimensions of the box is divisible by n.
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