Hello there, everyone!
Today I want to take you through a well-known math problem which I enjoy calling “The Painted Cube Conundrum,” because it sounds like a Sherlock Holmes case. Also, who’s excited for Sherlock Season 4 in 2017? I know I am!
The problem usually goes something like this:
A 4x4x4 cube is painted red over its entire surface, and then cut into 1x1x1 blocks. How many of the 1x1x1 blocks will have 2 faces painted?
This is usually followed by people drawing a cube, drawing the divisions and trying to work out manually exactly how many little cubes will have 2 faces painted. This can be difficult, time-consuming, and you’re never quite sure if you’re right unless you physically fold (or maybe you can draw really well) a cube with 4x4x4 divisions drawn on it.
Eventually, however, you will get to an answer of 24 blocks with 2 faces painted. Well done!
However, when you come across another question like this, but this time maybe a 6x6x6 cube, you have to start all over, and draw and count from scratch. There is no rule that you can apply to solve the question quickly.
That’s where I come in.
This post is going to show you how to calculate, for a cube of any side length,
the number of little cubes with:
– 3 faces painted
– 2 faces painted
– 1 face painted
– No faces painted
Let’s start with the example of the 4x4x4 cube, and take it from there to a cube of any side length.
3 faces painted:
By far the easiest part to solve is how many little cubes will have 3 faces painted.
Examine the diagram on the right and you will see something interesting.
Only the blocks on the corner (the vertices) of the cube will have 3 faces painted. This means we only need to find out how many vertices the cube has.
If we take a 4x4x4 cube, we will see that it has 8 vertices. This means that 8 blocks will have 3 faces painted.
Interestingly enough, this will be true for any cube, no matter the side length. All cubes have 8 vertices, so if you ever come across a question when you are asked how many little cubes have 3 painted faces, the answer should be 8, assuming of course that the initial question follows the format of the question I posted above, with only a different side length.
ONE IMPORTANT EXCEPTION: when a cube has side length 1, it will only have one little cube, which means that little cube will have 6 faces painted. Therefore, this kind of question will only be asked with cubes of side length 2 or more.
Okay, so now that we know the answer to that particular section, let’s move on to how to find out how many blocks have 2 faces painted.
2 faces painted:
This is slightly more difficult than the previous section, but still doable. Again, I have attached a diagram which should help you to understand what I am attempting to explain.
As you can see on the diagram, only the blocks on the outer edges will have 2 faces painted. REMEMBER, this does not include the vertices, as they have 3 faces painted.
There are 12 edges on a cube (again, a useful fact to know) and in this case, there are 2 blocks on every edge with 2 faces painted.
12 x 2 = 24
Therefore, there are 24 blocks with 2 faces painted in a 4x4x4 cube.
Now let’s take the case of a cube where we don’t know the number of blocks in its side length. Let’s say the side length is n, where n could be any number equal to or greater than 2. We can’t let n equal 1 because, as I explained earlier, our problem wouldn’t work.
So n is some number greater than or equal to 2.
If you notice on the cube above, there are only 2 blocks per edge which are not painted on 2 sides, namely the vertices. For each of the 12 edges, all the blocks except 2 are painted. This means that if there are n blocks on a side of the cube, n-2 blocks are painted on 2 sides on each edge.
We have already said that there are 12 edges on a cube, so if one edge has n-2 blocks painted on 2 sides, the entire cube must have 12(n-2) blocks painted on 2 sides.
Let’s try this with our 4x4x4 block. Assume n=4 in this case.
12(4-2) = 12(2) = 24 blocks
We already determined in our calculations that there are 24 blocks painted on 2 sides in a 4x4x4 cube. This formula gave us the same answer. Therefore, we can say that this formula works. (Usually we would prove this with multiple examples, but this formula really does work, and I don’t think you want me to list ten more examples)
So for any cube with side length n, the number of blocks with 2 sides painted = 12(n-2).
Great! We’re getting somewhere now. Only 2 more sections to go, and you’ll all be able to solve any Painted Cube Conundrum you are faced with.
1 face painted:
Every block on the surface that is not on an edge has 1 painted face. In the case of the 4x4x4 cube, each face of the cube has 4 blocks with 1 painted face.
Since a cube has 6 faces, this means 6×4 blocks with only 1 painted face, so 24 blocks.
In the nxnxn cube, we can see that the blocks with one painted face form a square on each face, with side length of n-2. The area of this square is (n-2)2. This means that on each of the 6 faces, there are (n-2)2 blocks with 1 painted face.
On the entire cube, the number of little blocks with 1 painted face is 6(n-2)2.
Applying this to the 4x4x4 cube, this would be:
6(4-2)2 = 6(2)2 = 6(4) = 24.
As you can see, again the formula holds for the example we have already worked out.
Now all that’s left is to see how many little blocks there will be with NO faces painted:
NO faces painted:
I don’t have a diagram for this one, since that was rather difficult to depict. So I’m just going to explain this one. If you have any questions, let me know in the comments and I’ll do what I can to clear up any misunderstandings.
If you remove the outer blocks (the ones with paint on them), you will find a smaller cube underneath. In the case of the 4x4x4 cube, the smaller cube will be a 2x2x2 cube, which means there will be 8 blocks with no paint on them.
In the case of an n x n x n cube, removing the outer layer of little blocks (the ones with paint on them) leaves us with a cube of (n-2) x (n-2) x (n-2). All these blocks have no paint on them, therefore the number of blocks with no paint on them will be (n-2)³.
If we use the 4x4x4 cube as an example:
(4-2)³ = (2)³ = 8.
As we can see, this formula holds true to the observation made earlier.
To summarise all these rules:
3 faces painted = 8
2 faces painted = 12(n-2)
1 face painted = 6(n-2)²
NO faces painted = (n-2)³
And there you have it. You should be able to solve whatever cube is thrown at you simply by following these four rules.
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Also, I will be launching an online Fundamental Math course soon where I will personally be teaching you some concepts and how to apply these to different types of problems.
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