There are 6 glasses on a table with their open ends up. In a move, we turn over any 4 of them. What is the minimum number of moves required to have all glasses upside down? Initially, we let D be a glass that's facing down and U be one that's up. At the beginning, all are up. After the first move, we know that we'll have 4 down and 2 up. Suppose that the first 4 are all down. From here, it's not possible to have all glasses facing down after a single move. We can flip the 4 glasses in the middle to obtain this pattern. And finally, flip the glasses that are up to obtain down, down, down, down, down, down. Thus, our answer is B, 3.

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minimum moves

upside down