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 Cube the Cube (Posted on 2004-10-02)
Remember Square Divisions? This problem demonstrates the deconstruction of a square into smaller squares with integer-length sides.

Given a cube with edge length 60, can you find a deconstruction of the cube into smaller cubes (none of which are alike) with integer length sides (or prove it can't be done)?

 No Solution Yet Submitted by SilverKnight Rating: 2.6667 (3 votes)

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 probably no solution | Comment 4 of 12 |

I'm thinking no solution exists...

1.  The solution, if it exists, would have to consist of at least 8 sub-cubes, because there are 8 corners on the original cube, and no two of these can be part of the same sub-cube.

2.  In fact there must be more than 8 cubes in the solution, because: Assume there are exactly 8, one in each corner.  If corner A has edge x, then the adjacent corner B would have edge 60-x.  But so would another adjacent corner C.  B and C would be the same size, which isn't allowed.

3.  If there's a solution,  each face of the cube would display a distinct solution to the similar puzzle for dividing a 60 x 60 square into smaller integer-sided squares.  No two such faces would have the same size sub-square unless they shared an edge on the large cube, and then the only squares they'd share would be along that edge, in the same sequence.  Hmm.  this feels like it's getting close to the impossibility proof.  (a) Are there at least 6 distinct solutions to the 60x60 square? (b) Can it be easily proved that no 2 such solutions could share the same sequence of sub-squares along one edge?

4.  Some preliminary thoughts about the smallest possible cube in each corner.  For example, the 1x1x1 cube can't go in the corner because the 3 cubes immediately adjacent would all have to be larger, and they'd interfere with one another. This means that the smallest cube in any corner would have to be the 3-cube with the 2-cube and 1-cube immediately adjacent to it. A larger cube could be placed on the 3rd edge, but that creates some problematic gaps that can only be filled with smaller cubes that have already been used.  So, what is the smallest possible corner cube?

In any case, intuition is telling me it's highly unlikely that a solution exists.  We're going to quickly run out of cubes to satisfy what appears to be a rapidly growing set of constraints.

Edited on October 3, 2004, 1:49 am
 Posted by Ken Haley on 2004-10-03 01:46:41

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