# Inside Out

There are 27 small cubes in a 3 x 3 x 3 cube, 54 faces being visible at any one time. Is it possible to reorganise these cubes so that by dipping the large cube into a pot of paint three times you can colour every face of all of the smaller cubes?

## Problem

Imagine a (2x2x2) cube made from 8 smaller red cubes. Each small cube has 6 faces, making a total of 6 x 8 = 48 small faces.

On each side of the large cube you can see faces of 4 of the smaller cubes. By turning the large cube around you can see 24 small faces all together.

You dip the cube into a pot of yellow paint and wait for it to dry. Now you separate all the smaller cubes and you can see that 24 of the 48 small faces are yellow and 24 are still red.

This suggests that you can rearrange the cubes so that only red faces are on the outside and yellow on the inside. Then, by dipping the cube into a pot of green paint it seems reasonable to assume you will end up with 8 small cubes whose 48 faces are now all painted yellow or green (24 of each), none that are still red and none that have been painted twice. Is this true?

How about a 3 x 3 cube? This will have 27 smaller cubes, making a total of 162 faces, 6 x 9 = 54 being visible at any one time. This suggests you might need three pots of paint to colour all the faces of the smaller cubes without repetition and without having any red faces left.

Is this possible and how would you rearrange the cubes each time? Describe the system you are using.

How about a 4 x 4 x 4 cube and an n x n x n cube?

N.B. This problem is a 3D version of "On the Edge " which can be found on the site.

## Student Solutions

Andre's (Tudor Vianu School) started of by giving his thoughts on the problem. Many thanks for this Andre - I think it is an excellent attempt to unpick what is happening.

**For a 2 x 2 Cube**

#### For a 3 x 3 Cube

Now, to analyse further, for increasing numbers of small cubes, I will create a table, with columns indicating: the side of the big cube, the total numbers of small cubes, the total number of faces, the number of faces at the exterior, and the number of colours that should be used (in principle, obtained as a division of the total number of faces by the number of exterior faces):

Side of cube | Small cubes | Faces (total) | Faces (exterior) | Colours |

2 | 8 | 6 x 8 = 48 | 6 x 4 = 24 | 2 |

3 | 27 | 6x 27 = 162 | 6 x 9 = 54 | 3 |

4 | 64 | 6 x 64 = 384 | 6 x 16 = 96 | 4 |

n | n x n x n | 6 x n x n x n | 6 x n x n | n |

- there are 8 cubes with three coloured faces (vertices)
- there are 12 cubes with 2 coloured faces (centres of edges)
- there are 6 cubes with 1 coloured 1 face (centres of faces)
- there is 1 cube with no coloured faces (centre of big cube)

- the cube from the centre must go into one vertex, to maximise its coloured faces in the following step
- one cube from one vertex must go into the centre of the big cube
- the other cubes from the vertices must go into the centres of the faces (6), and one in a centre of the edges
- the cubes from the centres of the faces must go into the vertices

For the 3x3x3 cube, 2 of the corner slots and the middle slot can be used to swap 3 cubes around and colour then in the three goes. The rest of the 6 corner pieces are moved to 6 of the edge pieces, the 6 edges pieces are moved to the 6 centre slots and the 6 centre pieces are moved to the corners. Doing this after the second dipping as well will cover all 18 of these pieces in the three goes. Finally, the remaining 6 edge pieces can be left where they are, and simply rotated each time so that after the three dippings they are all covered. Thus a 3x3x3 cube can be totally coated.

For an n x n x n cube (if n is 6 or greater, or is 4), can be coloured as follows:

- We start of with 8 corner pieces, 12(n-2) edge pieces, 6(n-2)^2 centres, and (n-3)^3 middles
- 4n blocks are looped around spending 2 dips in the corners, and (n-2) dips in the middles. This leaves 12(n-2) edges, 6(n-2)^2 centres and n(n-2)(n-4) middles.
- 6n(n-2) gaps are kept for blocks spending 2 times at the edge, 2 times at the centre, and (n-4) times in the middle

If n=4 then all the blocks are covered. If n is greater than or equal to 6, then there are still some edge and middle pieces we haven't describe. Blocks loop around these n(n-2)(n-4) slots, spending 6 times at the centres, and (n-6) times at the edges.

The 'missing' case is n=5. Here instead of the second point we would use 52 of the centre slots and 13 of the middle slots to move around a total of 65 cubes in to cover them in the five dips. We would then use 2 of the edge slots, two of the centre slots, and one middle to colour 5 more blocks in the 5 dips. Finally we would use the 21 edge slots and 14 middle slots to cover a total of 35 blocks using a total of 5 dips (so each block spends two 3 times at an edge and 2 times in the middle).

So an n*n*n block can be covered in n dips of paint as expected.

## Teachers' Resources

This problem benefits from being able to handle real cubes using multilink or similar equipment.

Some sticky labels to identify which cubes have been painted and their locations may also prove beneficial.

Just trying to justify that in theory this should also be possible is a useful little proof.