Please solve the following question:

If a cube is cut into finite number of smaller cubes, prove that at least two of them must be of same size.

Results 1 to 12 of 12

- May 29th 2006, 06:53 PM #1

- May 29th 2006, 07:11 PM #2

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 9

- May 29th 2006, 07:32 PM #3

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 4

- May 29th 2006, 07:33 PM #4

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 9

- May 29th 2006, 08:13 PM #5

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 4

Originally Posted by**ThePerfectHacker**

means that if two of the smaller cubes were adjacent before disassembly

of the big cube then the sum of their sides must be <=6 otherwise they

cannot have fitted into the cube of side 6.

RonL

- May 30th 2006, 01:58 PM #6

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 11,548
- Thanks
- 539

- May 30th 2006, 02:12 PM #7

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 9

- May 30th 2006, 05:14 PM #8

- May 30th 2006, 06:41 PM #9

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 9

- May 30th 2006, 09:26 PM #10

- May 31st 2006, 01:41 PM #11

- Joined
- Nov 2005
- From
- New York City
- Posts
- 10,616
- Thanks
- 9

- May 31st 2006, 04:49 PM #12