Remember
Square Divisions? This problem demonstrates the deconstruction of a square into smaller squares with integerlength 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)?
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 20041203 08:13:03 