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I'm not sure exactly how to visualize this riddle. Can someone help me?. It goes as this:

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 begans to take one cube at a time and paints some of its faces and places 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.

What I've attempted to so was using this logic:

What I'm assuming is 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 convinces 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 it wouldn't matter?. Can someone help me here?.

The original source of this riddle comes from a book titled Hability and Logic and dates from an unknown publisher going back 2019. I don't know if this is critical in the solution but looks required for reference.

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  • $\begingroup$ It doesn't make any mention cosntraining the orientation in which the faces are painted. I would interpret the question as "how many ways are there, ignoring rotation, to paint a cube in 2 colors?" (for which this becomes a more textbook math problem) $\endgroup$ – phenomist Mar 21 at 2:08
  • $\begingroup$ Where does this puzzle come from? If you have a puzzle from another source, you must give the exact source it comes from. $\endgroup$ – Deusovi Mar 21 at 2:27
  • $\begingroup$ @phenomist I think it was intended that way but not explicitly mentioning this. $\endgroup$ – Chris Steinbeck Bell Mar 21 at 2:51
  • $\begingroup$ @ChrisSteinbeckBell Shouldn't it be 'the cubes in the box'? $\endgroup$ – UnidentifiedX Mar 22 at 2:15
  • $\begingroup$ @ChrisSteinbeckBell And isn't it possible to paint multiple sides? $\endgroup$ – UnidentifiedX Mar 22 at 2:16
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Since this is a puzzle, one could interpret the condition "different from the cube which was already in the box" to refer to just the first cube (subsequent cubes weren't already in the box when the "After doing this" part starts). In that case, you could have as many cubes as the box can hold, so long as they were all different from the first cube.

If you count the ones in the box at any time, the number is still bounded only by the relative sizes of box and cube. Just make sure the paints are applied in different shades / colours each time.

If you only permit a single colour, then you can have 10 cubes in the box:

  • 0 painted faces: 1 (call it C1)
  • 1 painted face: 1 (call it C2)
  • 2 painted faces: 2 (C3: opposite faces painted, or C4: adjacent faces painted)
  • 3 painted faces: 2 (C5: C3 + any other face; C6: C4 + any that shares a vertex (if it doesn't share a vertex with the other 2 painted faces, it becomes C5))
  • 4, 5 and 6 painted faces are the inverse of 2, 1 and 0 painted faces, respectively, giving us 4: C7 & C8, 5: C9, 6:C10
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First, we find the number of ways possible to paint one side of the cube: 1, because the rest will repeat

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For two sides: 2

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For three sides, there is 2

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For four sides, there is 2

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For five, there is only 1.

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For all sides, I'm not sure if this is needed for the final solution but it has only 1 way.

Thus, when you add $1+2+2+2+1+1$, the answer will be 9.

You might need to try and visualise the rest of the faces because that is the best that I can do

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  • $\begingroup$ I consulted with the original source and the alternatives for this question are $7$, $10$, $9$ or $11$. Which among them could be the answer? $\endgroup$ – Chris Steinbeck Bell Mar 22 at 2:56
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    $\begingroup$ Your last edit is wrong. It was correct before as there really are 2 ways of colouring 3 faces of a cube - three faces adjacent to a corner, or two opposite faces plus one of the other faces. $\endgroup$ – Jaap Scherphuis Mar 22 at 12:03
  • $\begingroup$ @JaapScherphuis Yes, thanks for pointing out that :D $\endgroup$ – UnidentifiedX Mar 22 at 12:42

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