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if you go from Solved to Solves, you turn 210 times which equals 105 times R U. You would need to Pre-AUF (just for this R U case) for it to have an odd number of moves

I am not sure what you mean.
Doing R U 63 times (126 Moves) leaves me with five unsolved Corners. Doing R U 105 times (210 Moves) leaves me with a solved cube

I am not sure what you mean.
Doing R U 63 times (126 Moves) leaves me with five unsolved Corners. Doing R U 105 times (210 Moves) leaves me with a solved cube

I am thinking about buying the Gan Air SM but wonder if I should wait for the next 3x3 that Gan comes out with. I will buy the SM if it is a long time until they make a new 3x3. Im going to a competition in April and want a new cube by then, ether its the SM or a newer 3x3 by Gan. Does anyone know when Gan is releasing a new 3x3?

Why don't you generate twenty random-state scrambles and count how many of them have at least one pair?

(There are theoretical justifications as to why the probability "should" be close to 1−1/e, and my own empirical data agrees with this very well. Roux users should be very familiar with how free pairs show up in most solves.)

RUD and RUL overlap more than RUF due to RUF being able to change edges.

A more general answer:
If there's a solved 2x2x3 in DL and edges are oriented, then that case can be solved with any of RUD, RUL or RUF. However if there are any misoriented edges (here, "misoriented" means that it can't be solved any combination of R or U moves) you will need F moves to solve it.

In the specific case of PLL:
Any PLL (in fact, any ZBLL) can be solved with RUF or RUD or RUL, and each case also requires at most two F/L/D moves.

If you disassemble and reassemble a 3x3 cube, there are 519 024 039 293 878 272 000 combinations. But, since not all of these combinations are possible when turning the cube normally, you have to divide that number by 12 to get 43 252 003 274 489 856 000. If the 3x3 "divider" is 12, what is it on 4x4 (when ignoring internal pieces and using fixed centers)? What about 5x5?

If you disassemble and reassemble a 3x3 cube, there are 519 024 039 293 878 272 000 combinations. But, since not all of these combinations are possible when turning the cube normally, you have to divide that number by 12 to get 43 252 003 274 489 856 000. If the 3x3 "divider" is 12, what is it on 4x4 (when ignoring internal pieces and using fixed centers)? What about 5x5?

12 on all odd cubes (other than 1×1×1, of course) and 3 on all even cubes.

The "dividers" come from constraints on how the pieces can be placed if you want the puzzle to be solvable. On big cubes, the centres can be arbitrarily permuted independently of the other pieces (note: not true for supercubes), as can the wings (note: also not true for supercubes), so the only types of pieces left to consider are the corners and the midges.

Corner orientation contributes a factor of 3, always. Midge orientation contributes a factor of 2, but only for odd cubes. There's a further factor of 2 for odd cubes because the permutation parity of the corners and midges (taken together) must be even. Without the stipulation of having fixed centres, there would also be an additional factor of 24 for even cubes for how the cube is rotated.