Olympiad problem with cubes
 

Show that a cube can be dissected into n cubes for n >= 55

Re: Olympiad problem with cubes
 

<font color="white">It's obvious that a cube can be dissected into n^3 equal cubes. It's also obvious that n^3 of these equal cubes, aligned correctly, can be combined into one cube. Inductively, if 55-61 exist, then breaking any cube into 8 subcubes will generate all further numbers.

So the operations +7, +26, +63, +124, and +215 always exist, and the operations -7, -26, -63, etc. conditionally exist.

55: Split into a 3x3x3 cube (27), split 4 into 2x2x2 (+28)

56: Split into a 6x6x6 cube (216), condense 4 sets of 27 cubes into 3x3x3 cubes (-104 = 112) (leaving a 6x6x3 set of cubes on the bottom and 4 big 3x3x3 cubes on top), then condense 8 sets of 8 cubes (-56 = 56)

57: 64 - 7

58: 216 - 5*26 -4*7

59: 64 - 26 + 7*3

60: 27 + 26 + 7

61: 27 + 4*26 (blow up a 2x2 square on one side) - 10*7 (9 from the subsubcubes and 1 from the subcubes)

There's probably a proof based on a +3*26 - 11*7 being an increase of one cube, or someting close to that, but I'm not sure how to show 55's relevance in any clean way (my solution for 61 doesn't have room for another -7, and 216 - 3*26 - 11*7 also doesn't have room for another -7).</font>

Re: Olympiad problem with cubes
 

And it's not terribly difficult to show this for n&gt;=48.

Re: Olympiad problem with cubes
 

[ QUOTE ]
And it's not terribly difficult to show this for n&gt;=48.

[/ QUOTE ]

Do tell us the method for n = 54

Re: Olympiad problem with cubes
 

Everything between 48 and 53 is automatic from my n&gt;=55 solutions. I'm almost positive 54 requires condensing from a &gt; 6x6x6 starting cube (which if true, is probably why the problem is n&gt;=55, since they're exponentially harder to deal with).

Re: Olympiad problem with cubes
 

Using the "lingo" of the post by TC that was in white:

54 = 8^3 - 6 x 63 - 2 x 26 - 4 x 7

(Okay, so it's not THAT easy to see this geometrically!)

Also, looking at everything (mod 7),

53 is congruent to 39 and 20 = 27-7, so 39 = 20+20-1 is
possible
(if a cube can be dissected into x1 subcubes or x2 subcubes
it obviously can also be dissected into x1+x2-1 subcubes)

EDIT: much easier is 53 = 27+26

52 is congruent to 38 = 64-26
51 = 6^3 - 5 x 26 - 5 x 7 (geometrically ok)
50 is congruent to 1
49 = 6^3 - 4 x 26 - 9 x 7 (geometrically ok)
48 is congruent to 27

47 is not possible

Re: Olympiad problem with cubes
 

How do you show 47 is not possible?

Re: Olympiad problem with cubes
 

[ QUOTE ]
How do you show 47 is not possible?

[/ QUOTE ]

Trial and error.