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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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 Alternative proof Comment 12 of 12 |

Alternative proof of impossibility, please check if it is valid.

You cannot use a 1 unit cube in the corner, any other cube touching one of the sides of this 1 unit cube will "stick out" and create a gap with the original cube that cannot be filled.

You cannot use a 2 unit cube in the corner, any cube (other then the 1 unit cube) touching the second of the sides of this 2 unit cube will "stick out" and create a gap with the original cube that cannot be filled.

You cannot use a 3 unit cube in the corner, any cube (other then the 1 and the 2 unit cube) touching the third of the sides of this 3 unit cube will "stick out" and create a gap with the original cube that cannot be filled.

When you use a 4 unit cube in a corner, you now have a volume that has 7 original corners and 3 extra corners (Between the 4 unit cube and the original).  The problem of filling 8 corners just got worse:  You have now 10 corners to fill.

Each of these extra corners cannot be filled with a 1 unit cube in the corner, any other cube touching one of the free sides of this 1 unit cube will "stick out" and create a gap with the 4 unit/original cube that cannot be filled. Same reasoning as above goes for 2 and 3.  Follows that you cannot use a 4 unit cube in the corner, so you must use a 5 unit cube.

When you use a 5 unit cube in a corner, you now have a volume that has 7 original corners and 3 extra corners (Between the 5 unit cube and the original).  You have now 10 corners to fill.

Each of these extra corners cannot be filled with a 1 unit cube in the corner, any other cube touching one of the free sides of this 1 unit cube will "stick out" and create a gap with the 5 unit/original cube that cannot be filled. Same reasoning as above goes for 2 and 3, you also proved already that  a 4 unit cube cannot be used in a corner.  Follows that you cannot use a 5 unit cube in the corner, so you must use a 6 unit cube.

Same reasoning as just above for 6,7,,... proves that you cannot use any cube in a corner except for the 60 unit cube.

 Posted by Hugo on 2004-12-03 08:13:03

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