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.

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Concepts:optimization pigeonhole principle

Difficulty rating: 1090

19.

There are $120$ 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?

$30$

$40$

$41$

$60$

$119$

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 occupied-empty-empty, filling the middle seat of every group of three.

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

Thus, the correct answer is B .

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