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How Many Cube Faces Were Painted?

Date: 02/23/2003 at 05:23:29
From: Jesse
Subject: Faces painted on a cube

Some unit cubes are put together to form a larger cube. Some of the 
larger cube's faces are painted and then it is taken apart. 45 small 
cubes are found to have no paint on them. How many faces of the large 
cube were painted?

I know what to do if all the faces were painted but I have no idea if 
only SOME of the faces were painted.

Date: 02/23/2003 at 07:12:58
From: Doctor Jacques
Subject: Re: Faces painted on a cube

Hi Jesse,

This is a beautiful problem...

Let n be the size of the edge of the cube.

Consider the "inner" cube, i.e. the cube formed by the small cubes 
that are not on the faces on the large cube. The size of the edge of 
the inner cube is (n-2), and it contains (n-2)^3 small cubes.

As none of the cubes of the inner cube can receive paint, there are 
at most 45 of them (otherwise, we would find more than 45 unpainted 

This means that:

  (n-2)^3 <= 45

from which we deduce that n-2 is at most 3, i.e. n <= 5.

On the other hand, as we found 45 small cubes, the large cube 
contained at least 45 cubes, i.e.:

  n^3 >= 45

which gives n >= 4.

We can conclude that n is 4 or 5.

Assume first that n = 4. We have 2^3 = 8 interior cubes, and there 
must be 45-8 = 37 exterior unpainted cubes.

The number of exterior cubes is:

  4^3 - 2^3 = 56

As 37 of these 56 cubes are not painted, there are 56-37 = 19 painted 

Note that, when counting painted cubes, we cannot simply add the 
number of cubes on each face: if we paint two adjacent faces, the 
cubes on the common edge will be painted twice, but we must count 
them only once.

As each face contains 4^2 = 16 cubes, we must paint at least two faces 
to get 19 painted cubes. The first face will give 16 cubes; the other 
face will give 16 other cubes if it is not adjacent to the first one, 
and 12 other cubes if it is (because there are 4 cubes on the common 
edge). In total, we will have at least 28 cubes, and this is already 
more than 19.

We can conclude that n is not 4, and thus n = 5.

By repeating the calculations above, you will find that there are:

 125 cubes in total
 27 interior cubes
 98 exterior cubes
 18 exterior unpainted cubes
 80 exterior painted cubes.

As each face contains 25 cubes, painting a face can produce _at most_ 
25 new painted cubes (it can produce less, if some faces are adjacent, 
as explained above).

As we need 80 painted cubes, we need to paint at least 4 faces.

If we paint 5 faces, only one face will remain unpainted, and the only 
unpainted cubes will be those of the inner square of that face (those 
adjacent to an edge will have paint on them from another face). This 
will only leave 3^2 = 9 unpainted exterior cubes, which is not enough.

This means that there are 4 painted faces. There are still two 
possiblities: the two unpainted faces can be opposite or adjacent. 
I'll let you count the cubes in each case and find out which is the 
correct one.

Please feel free to write back if you want to discuss this further.

- Doctor Jacques, The Math Forum 
Associated Topics:
High School Polyhedra
High School Puzzles
Middle School Polyhedra
Middle School Puzzles

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