Problem 9 · AMC 8 Stretch
Stretch
Counting & Probability
Logic & Word Problems
pigeonholelogical-reasoningsymmetry
A round table has 5 chairs, and a name card is taped to the table in front of each chair. Five friends sit down without looking, and it turns out nobody is in front of their own card. Show that you can spin the table to a new position where at least 2 friends end up in front of their own cards.
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Answer: some spin makes at least 2 friends correct
Show hints
Hint 1 of 4
Spinning the table one chair at a time moves every card forward one seat. There are 4 'new' spin positions (after 1, 2, 3, or 4 clicks); the starting position has nobody correct.
Still stuck? Show hint 2 →
Hint 2 of 4
Pick one friend. As the table makes its 4 clicks, that friend's own card passes by their seat exactly once. So exactly one spin position is 'correct' for that friend.
Still stuck? Show hint 3 →
Hint 3 of 4
Every friend has exactly one correct spin position among the 4. Make the 4 spin positions your boxes, and assign each friend to their correct one.
Show solution
Approach: Pigeonhole — 5 friends share only 4 nonzero spin positions
- Spin the table one click at a time. There are 4 'new' positions (1, 2, 3, or 4 clicks); the no-spin start has nobody correct, by assumption.
- Focus on one friend. As the cards parade past that friend's seat over the 4 clicks, the friend's own card lines up with the seat exactly once (not at the start, so during one of the 4 clicks). So each friend has exactly one correct spin position.
- Make the 4 spin positions our boxes and put each of the 5 friends into the box for their one correct spin. Since \(5 > 4\), two friends land in the same box.
- At that single spin position, both of those friends sit in front of their own cards — so some spin makes at least \(2\) friends correct.
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