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

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==Solution==
 
==Solution==
  
If there are <math>120/2=60</math> seats occupied, then it is possible to get an occupation of the seats where no person sits next to someone else.  However, the only way is through alternating seats.
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Let <math>p</math> be a person seated and <math>o</math> is an empty seat
  
Thus, if another person joins, two people must sit next to each other, so <math>\boxed{\text{D}}</math>.
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The pattern of seating that results in the fewest occupied seats is <math>\text{opoopoopoo...po}</math>.
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We can group the seats in 3s like this: <math>\text{opo opo opo ... opo}.</math>
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There are a total of <math>40=\boxed{B}</math> groups
  
 
==See Also==
 
==See Also==

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions