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)?

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**