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Cube the Cube (Posted on 2004-10-02) Difficulty: 5 of 5
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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Solution Solution | Comment 9 of 12 |
Tweaked a few minor errors in the proof:

It is impossible to decompose a 60x60x60 cube into smaller cubes of distinct integral side lengths.  To see this, we first define some terms:

Define a cube to be big if its side length is greater than 21.
Define a cube to be very big if its side length is greater than 31.

Detti correctly notes that a decomposition cannot contain any very big cubes.  Another way of seeing is this is to note that a decomposition whose largest cube is a very big cube of length n cannot use any other cube of length greater than (60-n).  Then simply write a  computer program to verify that n^3 + (1^3+2^3+...(60-n)^3)<60^3 for 31<n<60.

I now claim that every decomposition must contain at least 9 big cubes.  To see this, note that a decomposition that has  fewer than 9 big cubes has total volume at most (1^3+2^3+...+28^3+29^3+31^3) - 22^3, which we can verify by computer to be less than 60^3.  This is because the decomposition has no very big cubes, and because the decomposition cannot contain both a 30x30x30 cube and a 31x31x31 cube (note that 30^3 is left out of the parenthized sum).

Finally, I now claim that it is impossible to fit 9 big cubes in a 60x60x60 cube.  To see this, note that we cannot fit nine identical 22x22x22 cubes into a 60x60x60 cube.  This last fact is rather obvious, but I'll give a rigorous proof for completeness:

Look at the bottom 60x60x22 region of the cube.  This contains parts of at most four of the 22x22x22 cubes, since only four 22x22 squares can fit into a 60x60 one.  It suffices to show that the remaining 60x60x38 region cannot fit five full 22x22x22 cubes.  Applying the same reasoning again, it suffices to show that a 60x60x16 region cannot fit a 22x22x22 cube, which is clear.

Edited on October 8, 2004, 5:02 pm
  Posted by David Shin on 2004-10-08 16:49:33

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