How can I find the least number of cubes which can be in a box?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

Mike has many identical cubes whose faces are white. He decides to take one of those cubes and places it inside of an empty box. After doing this, he begins to take one cube at a time and paints some of its faces and places it in the box. However, he keeps doing this only if the cube is different from the cube which was already in the box. Find the maximum number of cubes which can be in the box.

The alternatives given are:

$\begin{array}{ll}
1.&\textrm{7 cubes}\\
2.&\textrm{10 cubes}\\
3.&\textrm{9 cubes}\\
4.&\textrm{11 cubes}\\
\end{array}$

I'm not sure exactly how to visualize correctly this problem. But I'm assuming that the maximum will be perhaps $7$? As each block will have its color "to remain white" as it is slowly being filled with green color. There's a maximum of $6$ faces in a cube. So adding this to the existing white block inside the box, it would be $7$. Am I right with this conclusion? If possible, can someone add some drawing to justify this?

The part where it makes me feel not very convinced is that I'm assuming that he paints the faces contiguously to each other. But what if he skips one face, how can I account for it? Or wouldn't it matter? Can someone help me here?
 

skeeter

Math Team
Jul 2011
3,355
1,848
Texas
as I interpret the problem

ways to paint 0 faces ... 1
ways to paint 1 face ... 1
ways to paint 2 faces ... 2 (2 adjacent faces, or 2 opposite faces)
ways to paint 3 faces ... 2 (2 opposite & 1 adjacent, or all 3 adjacent)
ways to paint 4 faces ... 2 (complement of painting 2 faces)
ways to paint 5 faces ... 1 (complement of painting 1 face)
ways to paint 6 faces ... 1 (all 6 faces painted)

10 total
 
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