# Breaking a larger cube into a smaller one

I would appreciate ideas on solving the following problem:

A solid cube of side 6 is first painted pink and then cut into smaller cubes of side 2.How many of smaller cubes have paint on exactly 2 side ?

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There is no "trick" to it. It is just visualization. – picakhu Jul 25 '12 at 13:37

12, the answer is 12. Simple visualization will yield you this...

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Imagine a Rubik's cube. There are 6 corners, each 3 colored, then 8 edges that have only 2 colors, and finally there are 6 centre faces that have only one color. So if its all painted the same color then obviously you'll have 8 such cubies...

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Now that you have solved the problem, you can generalize. The outside of an $n\times n \times n$ cube is painted green, a sustainable colour. Then the cube is divided into $n^3$ $1\times 1\times 1$ cubelets.

How many cubelets have two coloured sides? A sketch shows that these are the cubelets that have exactly one edge along an edge of the big cube.

The big cube has $12$ edges. For any given such edge, there are $n-2$ cubelets that have exactly one edge along an edge of the original cube. Thus we get a total of $12(n-2)$ cubelets with two green faces.

Note that only the $8$ corner cubelets have $3$ green faces.

For $1$ green face, each of the $6$ faces of the big cube gives us $(n-2)^2$ cubelets with $1$ green face, for a total of $6(n-2)^2$.

Finally, how many totally uncoloured cubelets? The $(n-2)^3$ "inside" ones.

Let's check. There are $n^3$ cubelets. Of these, $(n-2)^3$ are uncoloured, $6(n-2)^2$ have one green face, $12(n-2)$ have $2$ green faces, and $8$ have $3$ green faces. So we should have $$n^3=(n-2)^3+6(n-2)^2+ 12(n-2)+8.$$ Expanding shows this is true. There is an easier way, by using the formula for $(x+y)^3$, with $x=n-2$ and $y=2$.

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Take a unit line segment and paint the two endpoints. Then cut it into three parts. This produces a piece on the left end with a painted endpoint and an unpainted endpoint ($Pu$), a piece in the middle with two unpainted endpoints ($uu$), and a piece on the right end with an unpainted and a painted endpoint ($uP$). We can write this as $Pu+uu+uP$, where $P$ means a boundary that has paint and $u$ means a a boundary that is unpainted.

We want to know how many parts of $(Pu+uu+uP)^3$ have exactly two painted boundaries. We can rewrite $Pu+uu+uP$ as $2Pu+u^2$ and expand $(2Pu+u^2)^3$ using the binomial theorem:

$$\begin{eqnarray} (2Pu+u^2)^3 & = & (2Pu)^3 + 3(2Pu)^2(u^2) + 3(2Pu)(u^2)^2 + (u^2)^3\\ &=&8P^3u^3 + 12P^2u^4 + 6Pu^5 + u^6 \end{eqnarray}$$

Thus there are 8 parts with three sides painted and three unpainted ($P^3u^3$), 12 parts with two sides painted and four unpainted ($P^2u^4$), and so on; the answer is 12.

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This seems to be the best way for generalization. Very creative too! – karvens Oct 12 '14 at 14:37

Here's my hint: Stare at that for a sufficiently long time, and the solution will come up.

and if that doesn't help, try this:

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A couple of hints:

The only ones which have three painted faces must be at the corners (vertices). How many vertices does a cube have?

The only ones which get one painted face must be at the centre of a face of the big cube. How many faces does a cube have?

Incidentally, when you have solved that one, I would recommend having a go at some bigger cubes, and then see if you can spot any patterns. Generalise - and then see if you can prove your answers must be correct.

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The two-faced cubes are all "edge-cubes". There is one on each edge of the cube. The corner cubes have 3 painted faces, the middle of face cubes have one, and the interior cube is unpainted.

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I would start by counting how many of the small cubes there are all together. Then try to find which ones have 2 sides painted.

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Total of 27 smaller cubes – Rajeshwar Jul 25 '12 at 14:00