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

(Solution)
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==Solution==
 
==Solution==
  
Every third seat is occupied resulting in 40 seats in addition to seating the first chair thus <math>\boxed{41}</math>
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p is a person seated, 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
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opo opo opo ...opo
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 +
there are a total of <math>\boxed{40}</math> groups
 
==See Also==
 
==See Also==
  
 
{{AJHSME box|year=1990|num-b=18|num-a=20}}
 
{{AJHSME box|year=1990|num-b=18|num-a=20}}
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 20:54, 14 September 2011

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

p is a person seated, 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 opo opo opo ...opo

there are a total of $\boxed{40}$ 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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