1990 AMC 8 Problem 19
Attempt Problem 19 of the 1990 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
19.
There are seats in a row. What is the fewest number of seats that must be occupied so the next person to be seated must sit next to someone?
Answer: B
Small hint:
To force the next person to sit adjacent to someone, no empty seat may have both neighbors empty
Big hint:
Seat people so there is exactly one empty seat between occupied ones: occupy every third seat
Solution:
To force the next person next to someone, every empty seat must be adjacent to an occupied one. Each occupied seat can account for at most itself and its two neighboring seats, so at least seats must be occupied.
This bound is attainable: use a repeating pattern of empty-occupied-empty, filling the middle seat of every group of three. This uses exactly occupied seats.
Thus, the correct answer is B .
Problem 19 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
- 1995 AMC 8
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- 1999 AMC 8
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