1. 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.

2. 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.
$\displaystyle 3^3+4^3+5^3=6^3$

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

4. 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.

5. 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

6. Hello, ThePerfectHacker!

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.

7. I see what you are saying.

8. 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

9. Can I rephraze problem az,
$\displaystyle \sum_{k=1}^n x_k^3=y^3$
Where, $\displaystyle x_i< x_j$ iff $\displaystyle i<j$.

10. Originally Posted by ThePerfectHacker
Can I rephraze problem az,
$\displaystyle \sum_{k=1}^n x_k^3=y^3$
Where, $\displaystyle x_i< x_j$ iff $\displaystyle i<j$.
Keep in mind that sum of some x's should be y(edge of cube)

11. Originally Posted by malaygoel
Keep in mind that sum of some x's should be y(edge of cube)
$\displaystyle x_1+x_2+...+x_n=y$
With,
$\displaystyle (x_1)^3+(x_2)^3+...+(x_n)^3=y^3$

12. Originally Posted by ThePerfectHacker
$\displaystyle x_1+x_2+...+x_n=y$
With,
$\displaystyle (x_1)^3+(x_2)^3+...+(x_n)^3=y^3$
I think you have made the wrong equations