Difference between revisions of "1990 AJHSME Problems/Problem 19"
(→Solution) |
Pieslinger (talk | contribs) |
||
Line 7: | Line 7: | ||
==Solution== | ==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 <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 19:54, 14 September 2011
Problem
There are 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?
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 groups
See Also
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 |