Difference between revisions of "1990 AJHSME Problems/Problem 19"

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Latest revision as of 00:59, 25 November 2020

Problem

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?

$\text{(A)}\ 30 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 41 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 119$

Solution

Let $p$ be a person seated and $o$ is an empty seat

The pattern of seating that results in the fewest occupied seats is $\text{opoopoopoo...po}$. We can group the seats in 3s like this: $\text{opo opo opo ... opo}.$

There are a total of $40=\boxed{B}$ groups

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AJHSME/AMC 8 Problems and Solutions
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