Seven glasses, flip exactly four at a time

Question

Seven drinking glasses stand on a table, all right-side up. On each move you must choose exactly four of the glasses (any four you like) and turn each of those four upside down (or right-side up again, if it was already flipped). You may make as many moves as you want.

Can you ever reach the position where all seven glasses are upside down? If not, explain why it is impossible.

Accepted Answer

It is impossible, and a parity count shows why in one step. Track how many glasses point up. Call this number the "up-count." It starts at 7, which is odd. The goal, all seven upside down, has an up-count of 0, which is even. So the real question is: can a move ever flip the parity (odd/even) of the up-count? A move of four flips changes the up-count by an even number. Suppose that among the four glasses you flip, k of them were pointing up and the other 4 - k were pointing down. After the flips, those k turn down and those 4 - k turn up. The change in the up-count is (4 - k) - k = 4 - 2k, wh…

Seven glasses, flip exactly four at a time

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