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November 1st, 2018, 12:16 PM   #1
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Cube, induction, block

The cube S with side 2 consists of 8 unit cubes.

Define block a figure that is formed from a cube S as a result of the removal of one unit cube.

Decide if the cube with dimensions 2^n⋅2^n⋅2^n from which one cube has been removed can be built by blocks.


Since a block is formed by 7 unit cubes, we just need to prove that

7|2^(3n)−1=8^n−1

So, this is true

but my teacher told me "7|8n−1 is a necessary condition for a solution, but I don't think on its own it is a sufficient condition."

So... What's next? I don't know... Help me please

Last edited by skipjack; November 2nd, 2018 at 06:51 AM.
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November 2nd, 2018, 06:53 AM   #2
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For n > 1, I suspect it can't be built from undivided blocks.
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November 2nd, 2018, 08:20 AM   #3
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Quote:
Originally Posted by estera View Post
The cube S with side 2 consists of 8 unit cubes.

Define block a figure that is formed from a cube S as a result of the removal of one unit cube.

Decide if the cube with dimensions 2^n⋅2^n⋅2^n from which one cube has been removed can be built by blocks.


Since a block is formed by 7 unit cubes, we just need to prove that

7|2^(3n)−1=8^n−1

So, this is true

but my teacher told me "7|8n−1 is a necessary condition for a solution, but I don't think on its own it is a sufficient condition."

So... What's next? I don't know... Help me please
It is not sufficient to show that an integer number of "blocks" contains the same number of unit cubes as a cube with a side-length of 2^n and 1 unit cube removed. You have to show that they can be combined into a particular shape.

It is of course trivially true that such a defective cube with a side-length of 2^1 can be formed from 1 block. So that gives your base case for an induction. But you have to show how the 2^(k+1) case can be deduced from the 2^k case. And that is not just a matter of number of unit blocks. It involves a description of how to create a particular shape.

When I consider trying a proof by induction, I like to start by considering the cases for n = 2 and n = 3. So consider the case where n = 2. That would be a cube (missing one unit cube somewhere) with dimensions 4x4x4. That will contain 63 unit cubes. And 9 "blocks" contain 63 unit cubes. But how do I fit those 9 "blocks" together to form the defective cube? The issue is how you fit the 9 blocks together so that 8 of the "holes" disappear while keeping the dimensions 4x4x4.

Figure that out, and then figure out how to generalize that process. If you cannot figure it out for n = 2, then try to prove it is impossible. After all, you were not asked to prove that it is possible.

EDIT: Of course, if skipjack's intuition is correct, it might be easier to start by trying to show why it is impossible. This might have something to do with what are the prime factors of the divisors you found.

Last edited by JeffM1; November 2nd, 2018 at 08:25 AM.
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