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

(Solution)
(Solution)
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==Solution==
 
==Solution==
  
p is a person seated, o is an empty seat
+
Let <math>p</math> be a person seated and <math>o</math> is an empty seat
  
The pattern of seating that results in the fewest occupied seats is opoopoopoo...po
+
The pattern of seating that results in the fewest occupied seats is <math>opoopoopoo...po</math>.
we can group the seats in 3s
+
We can group the seats in 3s like this: <math>opo opo opo ... opo.</math>
opo opo opo ... opo
 
  
there are a total of <math>\boxed{B}</math> groups
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There are a total of <math>40=\boxed{B}</math> groups
  
 
==See Also==
 
==See Also==

Revision as of 00:58, 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 $opoopoopoo...po$. We can group the seats in 3s like this: $opo opo opo ... opo.$

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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