Painted faces on a cube

Question

Here's another challenge I used to give to my students:

Let's begin with a bunch of little white cubes assembled into a big white cube. All the little white cubes are equal.
Then I decide to paint some of the faces of the big cube blue. Afterwards I break the big cube apart into the smaller ones.
Only $24$ of the smaller cubes remain completely white.

How many cubes formed the big one? How many big faces did I paint?

Usually I use this problem to show how much we can deduce with so little information.

Answer 1

It hinges on the following insight:

Don't break the cube apart but keep the small cubes packed together. If you paint some faces of the cube, and remove the painted layers, you have a cuboid where each dimension is at zero, one or two units shorter than the original cube.

That means that the $24$ unpainted cubes

form a three-dimensional cuboid with dimensions that differ by at most two. The only way to factor $24$ like that is as $24=2\times3\times4$. Therefore you started with a $4\times4\times4$ cube and removed three layers, two of them opposite each other.

Answer 2

In general, you can paint 6 sides of an N×N×N cube in 10 ways:

* 0 sides
* 1 side
* 2 adjacent sides
* 2 opposite sides
* 3 adjacent sides (a corner)
* 3 sides, two of them opposite (a band)
* 4 sides except 2 opposite
* 4 sides except 2 adjacent
* 5 sides
* 6 sides

which leaves you with these numbers of unpainted sub-cubes, respectively:

* N×N×N
* N×N×(N–1)
* N×(N–1)×(N–1)
* N×N×(N–2)
* (N–1)×(N–1)×(N–1)
* N×(N–1)×(N–2)
* N×(N–2)×(N–2)
* (N–1)×(N–1)×(N–2)
* (N–1)×(N–2)×(N–2)
* (N–2)×(N–2)×(N–2)

Given a number of unpainted cubes,

one needs to find a factorisation that fits one of those expressions.

As Jaap Scherphuis shows, the value of 24 fits just one, which yields only 1 answer.

Answer 3

My approach was to...

start off by bounding the possible dimensions of the big cube, and then find a valid permutation of face-painting from that subset. But after bounding the size, finding a permutation became unnecessary.

To determine...

a maximum bound: We can be certain that the inner cube of dimensions $(N-2)$ will remain untouched no matter how the big cube is painted, so that must hold an equal or fewer number of small cubes than how many are left untouched. $(N-2)^3 \leq 24$. Since $3^3$ is already $27$,

$N<5$ is a maximum bound.

Then to determine...

a minimum bound: If at least one face of the big cube is painted, then the largest number of untouched unit cubes possible is the total number of cubes, minus one face of cubes. $N^3-N^2 \geq 24$. Since $3^3-3^2$ is only $18$,

$N>3$ is a minimum bound.

Therefore...

$N$ must be greater than $3$ and less than $5$, so $4$ is the only possible answer. It is not necessary to determine exactly what pattern of sides were painted to be certain of the answer.

To address David G's point below, to round out the answer:

With a 4x4x4 cube, there are 56 outer cubes; 40/16 paint/blank. The most cubes you can paint with a face is 16, so you need more than 2 faces. meanwhile 4 faces painted leaves only 8 or 10 cubes blank. So 3 faces must be painted. Finding the permutation is still unnecessary for the total answer. But to be specific, 16 for one face, +12 for each single-adjacent space. So 3 faces painted in a U-shape on a 4x4x4 cube is the full description of the cube.

Answer 4

You can represent the unpainted volume like this:

$(n-x_1-x_2)(n-y_1-y_2)(n-z_1-z_2) = 24$
where $x_1,x_2,y_1,...$ are either $0$ or $1$, representing whether the face was painted.

This means all we need to do is

Factorise $24$ into 3 factors such that any two factors are at most $2$ apart

We know that

The prime factorisation of $24$ is $2\cdot 2\cdot 2\cdot 3$

Which means you have the following combinations:

• $2\cdot 3\cdot 4$
• $2\cdot 2\cdot 6$

Only one of which works, meaning that:

$(n-x_1-x_2)(n-y_1-y_2)(n-z_1-z_2)=2\cdot 3\cdot 4\\\therefore(n-1-1)(n-1-0)(n-0-0)=2\cdot 3\cdot 4$
Since $x_1+x_2+y_1+y_2+z_1+z_2$ sides are painted, we know that $3$ sides were painted.
Since $(n-0-0)=4$, we know that the original cube side length must have been $4$.

I haven't looked at the other answers so I hope I haven't done anything too similar to anyone else :)

Answer 5

Here is my way of looking at it:

I started off by representing situations with my favorite variable ‘x’ (I’m more of a math kind of person). For example, let x be the length of a side of the larger cube, so $x^3-x^2=24$ represents a situation where 1 face is painted blue, and those blue squares are removed ($x^3$ is the whole cube volume, so subtracting $x^2$ gives you the leftover volume after removing one face).

Then we move on to 2 blue faces...

$x^3-2x^2=24$ represents painting 2 blue faces that are opposite each other so the faces aren’t touching. And $x^3-2x^2+x=24$ represents 2 blue faces touching each other. The ‘plus x’ accounts for the fact that counting $x^2$ twice when the sides are touching means you are counting one edge two times; $2x^2$ tells you how many blue faces there are for the small cubes, so adding x will tell you how many cubes have any blue.

And so on... keeping in mind how many touching faces there are, we find that:

$x^3-3x^2+2x=24$ is the only equation out of all possible numbers of blue faces where a solution for x is a whole number ($x=4$). Remember that x represents the length of a side of the cube, so if the above equation is the only one with a whole number x, it’s the only one with a whole number side length and the only solution that works. The equation $x^3-3x+2x=24$ represents 3 blue sides, with 2 over laps (see coefficients). So since $x=4$, this means the whole cube has volume $4^3$ since x is AGAIN, the length of a side of the big cube.

Therefore...

You used 64 smaller cubes and painted 3 faces so that there are 2 edges where the faces touch (as in 2 of the faces are opposite each other).