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.

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- May 29th 2006, 06:53 PMmalaygoelNeed Help
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. - May 29th 2006, 07:11 PMThePerfectHackerQuote:

Originally Posted by**malaygoel**

$\displaystyle 3^3+4^3+5^3=6^3$ - May 29th 2006, 07:32 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

since if one cube were adjacent to one of the others at least one dimension

of the resulting solid must be 7 or greater.

RonL - May 29th 2006, 07:33 PMThePerfectHackerQuote:

Originally Posted by**CaptainBlack**

- May 29th 2006, 08:13 PMCaptainBlackQuote:

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 PMSoroban
Hello, ThePerfectHacker!

Quote:

I think I am missing something here.

Then you took the remaining volume, diced it into 91 one-inch cubes,

and*reassembled*them into a 3-inch cube and a 4-inch cube. - May 30th 2006, 02:12 PMThePerfectHacker
I see what you are saying.

- May 30th 2006, 05:14 PMmalaygoelQuote:

Originally Posted by**Soroban**

- May 30th 2006, 06:41 PMThePerfectHacker
Can I rephraze problem az,

$\displaystyle \sum_{k=1}^n x_k^3=y^3$

Where, $\displaystyle x_i< x_j$ iff $\displaystyle i<j$. - May 30th 2006, 09:26 PMmalaygoelQuote:

Originally Posted by**ThePerfectHacker**

- May 31st 2006, 01:41 PMThePerfectHackerQuote:

Originally Posted by**malaygoel**

With,

$\displaystyle (x_1)^3+(x_2)^3+...+(x_n)^3=y^3$ - May 31st 2006, 04:49 PMmalaygoelQuote:

Originally Posted by**ThePerfectHacker**