1990 AMC 8 Problem 19

Below is the professionally curated solution for Problem 19 of the 1990 AMC 8, 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).

Concepts:optimizationpigeonhole principle

Difficulty rating: 1090

19.

There are 120120 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?

3030

4040

4141

6060

119119

Solution:

To force the next person next to someone, every empty seat must be adjacent to an occupied one. The most efficient way is to seat people in a repeating pattern of empty-occupied-empty, filling the middle seat of every group of three.

With 120120 seats, this uses 1203=40\dfrac{120}{3} = 40 occupied seats.

Thus, the correct answer is B .

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