My mother is trying to solve a 2x2x2 Rubik’s cube. Sometimes she can get one face solid colored. Sometimes two faces and even three faces.

Is it possible that she could get 4 or 5 faces solid colored and still not have completely solved the cube?

  • $\begingroup$ Supose it is an ordinary cube, so each face is of one of 6 different colour. $\endgroup$
    – z100
    2 days ago

1 Answer 1


5 faces solved?

If 5 faces are solved, then the only remaining face colours are the sixth colour.
So it is impossible to have exactly 5 faces solved, but not the 6th.

4 faces solved?

Consider one face of a solved 2x2x2 cube.
Let's try to disrupt one adjacent face but leave the chosen face complete.

First, swap two adjacent corner pieces which connect that face.
But the effect will be to disrupt three adjacent faces.

Next, try to swap two opposite corner pieces.
But the effect will be to disrupt all four adjacent faces.

So it is impossible to have exactly 2 faces unsolved.

To explain further:
In larger cubes there are three kinds of piece: centre, edge and corner. But in a 2x2x2 cube there are only corner pieces. So every piece has three colours.

If you could rotate a single piece (but not possible) you change 3 faces.

If you swap 2 pieces such that one face stays the same colour, then each one affects 2 other faces – but they can't be the same 2 colours because each piece is unique. So it affects at least 3 faces.

  • $\begingroup$ Thanks for responding. I agree with you that the 5 color case is impossible but I don’t quite understand your 4 colors argument. $\endgroup$ Aug 5 at 22:42
  • 1
    $\begingroup$ @WillOctagonGibson I added more explanation. $\endgroup$ Aug 6 at 9:02
  • $\begingroup$ I am not convinced by the 4 faces argument. I think at the very least, a convincing argument would need to make a case-by-case analysis of the two possibilities "4 faces are correct, and the two remaining faces are adjacent faces" and "4 faces are correct, and the two remaining faces are opposite faces". $\endgroup$
    – Stef
    Aug 6 at 11:33
  • 1
    $\begingroup$ This answer uses the usual convention that a face is solved when all the pieces on that face are correctly placed. However, the OP's mother probably consider a face to be done when she can see only one color when looking at it (otherwise she couldn't get 3 faces solved but not the whole cube). So the white face would be "done" even if two diagonally opposed pieced are swapped, as long as they both show their white side on this face. $\endgroup$
    – Evargalo
    2 days ago
  • 1
    $\begingroup$ @Evargalo this reasoning supposes a face is solved when all 4 facets are the same colour - see its last paragraph. $\endgroup$ 2 days ago

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