4
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only ONE piece away from solved

I know that in a 3x3x3 this should never happen (unsolvable state), but is the same true in a 4x4x4?

(the rest of the cube is solved)

The following answer seems to say that everything about the 3x3x3 is true about NxNxN, except that the orientation isn’t fixed.

Why is a single-corner twist not a valid position on a Rubik's cube?

Does that mean that my orientation is wrong?

The cube came solved, and since this isn’t a speed cube, I don’t see how I could have accidentally twisted a corner.

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    $\begingroup$ I'm 99.5% sure it's impossible. If you can prove it's impossible on a 2x2x2 cube, that should be enough to extend to cubes of all sizes. $\endgroup$ – greenturtle3141 Aug 31 at 1:57
  • $\begingroup$ Sorry Joe, but you DID accidentally twist a corner. The corners behave the same on every nxnxn cube (n>1). $\endgroup$ – Christopher Mowla Sep 17 at 21:01
  • $\begingroup$ Yeah, @Christopher Mowla, I’m guessing someone was playing with it when it was mixed up, and that person twisted the corner, because I think I was careful with it. Nevertheless, I eventually looked up how to take it apart, and I fixed it. $\endgroup$ – Joe Sep 18 at 0:35
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It is impossible.

If you ignore all pieces but the corners, it is equivalent to a 2x2x2 cube. And it is impossible to turn one corner of a 2x2x2.

This is valid for any size, i.e. every NxNxN cube.

Credit goes to william122 who gave this answer first.

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Yes. It is impossible, because a 4x4 can be reduced to a 3x3, via reduction, and a corner would be exactly the same as a 3x3 corner.

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    $\begingroup$ You need to elaborate on this... A 4x4x4 does not have a fixed center, so it's not as easy as you think. $\endgroup$ – greenturtle3141 Aug 31 at 7:05
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It is impossible, but I'll try to explain a way to help understand.
Along each of the three axes of rotation there are four layers (I'm not sure if there's a more technical term) which can be freely rotated around their respective axis. Imagine shifting the centre gap so that the second layer becomes thicker and the third layer becomes thinner, like this:

 ___ ___ ___ ___      ___ ___ ___ ___      ___ ___ ___ ___
|   |   |   |   |    |   |   |   |   |    |   |   |   |   |
|___|___|___|___|    |___|___|___|___|    |___|___|___|___|
|   |   |   |   |    |   |   |   |   |    |   |   |   |   |
|___|___|___|___| -> |   |   |   |   | -> |   |   |   |   |
|   |   |   |   |    |___|___|___|___|    |   |   |   |   |
|___|___|___|___|    |___|___|___|___|    |___|___|___|___|
|   |   |   |   |    |   |   |   |   |    |   |   |   |   |
|___|___|___|___|    |___|___|___|___|    |___|___|___|___|

Imagine doing this 'all the way' so that layer 3 becomes impossibly thin. Now it is functionally identical to a 3 x 3 cube, and we have not removed any functionality from the initial 4 x 4 cube.
This demonstrates that the middle layers do not affect the corners at all.

Hope this helps.

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    $\begingroup$ As soon as you shift the cut away from the centre, it is no longer a functional cube. After a half turn of the face the shifted cut no longer lines up with itself so middle layer turns along one of the other axes are blocked, effectively bandaging the middle layers together. A better way to demonstrate that the middle layers do not affect the corners is to remove their stickers. $\endgroup$ – Jaap Scherphuis Oct 1 at 4:08

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