Flip three glasses right-side up in exactly three moves, turning two at a time
Three glasses stand in a row: the first and third are upside down, the middle one is right-side up. Each move must turn over exactly two glasses, not necessarily adjacent. Using exactly three moves, can you get all three glasses right-side up — and if not, why not?
Reveal the answer
It's impossible. Each move changes the count of upside-down glasses by an even number (+2, 0, or -2), so the parity of that count can never change. You start with two glasses upside down (an even number) and need to reach zero (also even) — which sounds fine, but reaching it takes an even number of 'net' flips, and three moves of two glasses each can't produce that in this configuration. The puzzle, a staple of Boris Kordemsky's collections of Russian recreational mathematics, is really a lesson in proving something can't be done.
— Boris Kordemsky, The Moscow Puzzles: 359 Mathematical Recreations — 1956; English translation 1972