1990 AJHSME Problems/Problem 19: Difference between revisions

Revision as of 19:54, 14 September 2011

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?

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

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

1990 AJHSME (Problems • Answer Key • Resources)
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

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