Problem 24 · AMC 8 Stretch
Stretch
Logic & Word Problems
logical-reasoningaccount-for-all-possibilities
Four glasses sit at the corners of a square table, each right-side-up or upside-down. You are blindfolded. On each turn you may feel any two glasses and flip none, one, or both. After each turn the table is spun randomly, so you lose track of corners. A bell rings the instant all four glasses face the same way. What is the fewest number of turns that GUARANTEES you can make the bell ring?
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Answer: 5 turns
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Hint 1 of 4
You can't control which corners you grab compared to last time (the spin scrambles them), but you CAN choose two ADJACENT (side-by-side) glasses or two DIAGONAL (across) glasses. Mix the two kinds.
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Hint 2 of 4
Turn 1: grab a diagonal pair and turn both right-side-up. Turn 2: grab an adjacent pair and turn both right-side-up. If no bell yet, exactly one glass is upside-down.
Still stuck? Show hint 3 →
Hint 3 of 4
Turn 3: grab a diagonal pair and flip BOTH. If no bell, the two down glasses are now adjacent (side by side).
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Approach: Mix diagonal and adjacent grabs to force all four to match
- Mix DIAGONAL and ADJACENT grabs so that no matter how the table spins, the glasses are forced toward all-the-same.
- Turn 1 (diagonal): set both right-side-up. Turn 2 (adjacent): set both right-side-up. If no bell, exactly one glass is down.
- Turn 3 (diagonal): flip both. If no bell, the two down glasses are now adjacent. Turn 4 (adjacent): flip both. If no bell, the two down glasses are now diagonal.
- Turn 5 (diagonal): flip both — now all four match and the bell rings. Every branch finishes by turn \(5\), and no faster plan is guaranteed, so the answer is \(5\) turns.
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