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5 turns (assuming you MUST turn 3 over each time)

DDDDDDDDDDD 11 down

UUUDDDDDDDD-1 3 up 8 down

UUDUUDDDDDD-2 4 up 7 down (turned one of the up back to down and 2 to up)

UUDUDUUDDDD-3 5 up 6 down (again turned one of the up back to down)

UUDUDUUUUUD-4 8 up 3 down

UUUUUUUUUUU-5 11 up

Obviously, there can be different orders here as well.

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5 turns (assuming you MUST turn 3 over each time)

DDDDDDDDDDD 11 down

UUUDDDDDDDD-1 3 up 8 down

UUDUUDDDDDD-2 4 up 7 down (turned one of the up back to down and 2 to up)

UUDUDUUDDDD-3 5 up 6 down (again turned one of the up back to down)

UUDUDUUUUUD-4 8 up 3 down

UUUUUUUUUUU-5 11 up

Obviously, there can be different orders here as well.

Yup statman is right.

With 3 moves, you can flip at most 9 cups

With 4 moves, you can flip all the cups, but would have needed to flip one back over.

With 5 moves, there will be 15 total flips...which is 11 plus an even number...so you can flip two cups 3 times and the rest once.

2 cups flipped 3 times + 9 cups flipped 1 time = 6+9 = 15 total cup flips

15 / 3 = 5 moves

So it will take at least 5 moves of inverting 3 at a time....but it is possible to do it with any odd number of moves greater than or equal to 5.

Edited by EventHorizon
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