A New Insight on the Famous Painted Block Problem

There is a famous, well-known problem in the world of “rich math tasks” that involves taking an n x n x n cube and painting the outside of it. Then you break apart the large cube into unit cubes (see image below cribbed from here for n=2 and n=3):

cubes

Notice that some of the unit cubes have 3 painted faces, some have 2 painted faces, some have 1 painted face, and some have 0 painted faces.

The standard question is: For an n x n x n cube, how many of the unit cubes have 3 painted faces, 2 painted faces, 1 painted face, and 0 painted faces.

[In case you aren’t sure what I mean, for a 3 x 3 x 3 cube, there are 8 unit cubes with 3 painted faces, 12 unit cubes with 2 painted faces, 6 unit cubes with 1 painted face, and 1 unit cube with 0 painted faces.]

Earlier this year, I worked with a middle school student on this question. It was great fun, and so many insights were had. This problem comes highly recommended!

Today we had some in house professional development, and a colleague/teacher shared the problem with us, but he presented an insight I had never seen before that was lovely and mindblowing.

Spoiler alert: I’m about to give some of the fun away. So only jump below / keep reading if you’re okay with some some spoilers.

First off, here are some manipulatives we used in our PD:

We used colors to represent the number of painted faces. So pink = three painted faces, blue = two painted faces, yellow = one painted face, green = zero painted faces.

And we, though standard arguments, saw that for an n x n x n cube

green cubes (zero painted faces) = $(n-2)^3$
yellow cubes (one painted face) = $6(n-2)^2$
blue cubes (two painted faces) = $12(n-2)$
pink cubes (three painted faces) = $8$

There are so many ways to come up with those individual formulas.

And clearly these add up to the total number of cubes. So we have:

$n^3=(n-2)^3+6(n-2)^2+12(n-2)+8$

Lovely. (Feel free to expand each term out to see that the left side and right side truly are equal.)

Here’s the insight my colleague shared with us.

If you look carefully at the right hand side, you can actually see $(n-2)^3+6(n-2)^2+12(n-2)+8$ as the same thing as $((n-2)+2)^3$ .

By applying the binomial expansion to $((n-2)+2)^3$ , each term in the expansion gives us the number of cubes which have 3 sides painted, 2 sides painted, 1 side painted, or 0 sides painted. Seriously, use the binomial expansion on $((n-2)+2)^3$ and see for yourself! WHAAAAA?

So here’s the question he posed to us, and I pose to you…

Why? Why the heck does each term in the binomial expansion for $((n-2)+2)^3$ give you the number of cubes with 3 sides, 2 sides, 1 side, and 0 sides painted?

I mean, we know it works algebraically. We can expand it out to see. But why, conceptually? Where do the $(n-2)$ and $2$ fit into everything?

When I finally figured it out, my mind was blown. So simple and elegant, yet so unintuitive for me. I’m not going to type out my insight here. Yet, anyway. Because I want to leave it for others to think through. But if there is interest, after a short while, I can definitely do an update with the conceptual insight!

7 comments

Seems to me that it’s because (n-2) is pieces not on the end, and (2) is pieces on the end. The expansion then gives you the pieces that are on the end of 0,1,2,3 faces of the cube, which correspond to the number of faces that have to be painted.

samjshah says:

April 20, 2016 at 2:00 pm

Lovely! Indeed this is right. I am having fun thinking about how I will explain this so someone who has never seen this sort of argument would understand it. Love it!

Reply
1. Madhav says:
  
  April 22, 2016 at 2:11 pm
  
  I am not sure I understand how Mr.K’s comment provides that insight. I am probably missing something. I realize the point about n-2 and 2, but should not his comment raise the question of *why* the coefficients of the binomial expansion give the number of pieces that appear on 0,1,2 and 3 faces of the cube?
  
  I was thinking more like this: Since n-2 and 2 are the terms, the binomial expands all terms to be a product of n-2 and 2. So what the coefficients give is the different combinations of cuboids possible given that each side of the cuboid can either have n-2 or 2 as length. That gives us 4 combinations of cuboids: (n-2)x(n-2)x(n-2), 2x2x2, (n-2)*(n-2)*2, 2*2*(n-2)
  
  I am not sure how to proceed from here to explain why the 3rd combination has all the 1 face painted pieces, while the 4th combination has all the 2 side painted pieces.
  
  I hope I made sense :-)

I think the part I left out is that you’ve got a distinction between “end” and “not end”, and the exponent on your binomial multiplication tells you how many dimensions you’re doing this in. Thus each successive term tells you how many of the next dimension you have an end piece for.

That means that not only does this work for cubes and then the degenerate linear case, but it will work for squares (in two dimensions) and hypercubes (in four) or even higher.

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