# Need Help

• May 29th 2006, 06:53 PM
malaygoel
Need Help
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 PM
ThePerfectHacker
Quote:

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

$3^3+4^3+5^3=6^3$
• May 29th 2006, 07:32 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
$3^3+4^3+5^3=6^3$

You cannot form a cube of side 6 from two cubes of sides 3 and 4 and 5.
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 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
You cannot form a cube of side 6 from two cubes of sides 3 and 4.

RonL

I think I am missing something here.
• May 29th 2006, 08:13 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I think I am missing something here.

We are talking about cutting a cube of side 6 into smaller cubes. That
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
Soroban
Hello, ThePerfectHacker!

Quote:

I think I am missing something here.
You took a six-inch cube and cut out a 5-inch-cube.

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 PM
ThePerfectHacker
I see what you are saying.
• May 30th 2006, 05:14 PM
malaygoel
Quote:

Originally Posted by Soroban
Hello, ThePerfectHacker!

You took a six-inch cube and cut out a 5-inch-cube.

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.

prove that there will be atleast two cubes of same size
• May 30th 2006, 06:41 PM
ThePerfectHacker
Can I rephraze problem az,
$\sum_{k=1}^n x_k^3=y^3$
Where, $x_i< x_j$ iff $i.
• May 30th 2006, 09:26 PM
malaygoel
Quote:

Originally Posted by ThePerfectHacker
Can I rephraze problem az,
$\sum_{k=1}^n x_k^3=y^3$
Where, $x_i< x_j$ iff $i.

Keep in mind that sum of some x's should be y(edge of cube)
• May 31st 2006, 01:41 PM
ThePerfectHacker
Quote:

Originally Posted by malaygoel
Keep in mind that sum of some x's should be y(edge of cube)

$x_1+x_2+...+x_n=y$
With,
$(x_1)^3+(x_2)^3+...+(x_n)^3=y^3$
• May 31st 2006, 04:49 PM
malaygoel
Quote:

Originally Posted by ThePerfectHacker
$x_1+x_2+...+x_n=y$
With,
$(x_1)^3+(x_2)^3+...+(x_n)^3=y^3$

I think you have made the wrong equations