Difference between revisions of "1991 AHSME Problems/Problem 15"
m |
(Added a solution with explanation) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 2: | Line 2: | ||
A circular table has 60 chairs around it. There are <math>N</math> people seated at this table in such a way that the next person seated must sit next to someone. What is the smallest possible value for <math>N</math>? | A circular table has 60 chairs around it. There are <math>N</math> people seated at this table in such a way that the next person seated must sit next to someone. What is the smallest possible value for <math>N</math>? | ||
+ | |||
+ | <math>\text{(A) } 15\quad | ||
+ | \text{(B) } 20\quad | ||
+ | \text{(C) } 30\quad | ||
+ | \text{(D) } 40\quad | ||
+ | \text{(E) } 58</math> | ||
+ | |||
== Solution == | == Solution == | ||
− | <math>\fbox{}</math> | + | <math>\fbox{B}</math> If we fill every third chair with a person, then the condition is satisfied, giving <math>N=20</math>. Decreasing <math>N</math> any further means there is at least one gap of <math>4</math>, so that the person can sit themselves in the middle (seat <math>2</math> of <math>4</math>) and not be next to anyone. Hence the minimum value of <math>N</math> is <math>20</math>. |
== See also == | == See also == |
Latest revision as of 16:38, 23 February 2018
Problem
A circular table has 60 chairs around it. There are people seated at this table in such a way that the next person seated must sit next to someone. What is the smallest possible value for ?
Solution
If we fill every third chair with a person, then the condition is satisfied, giving . Decreasing any further means there is at least one gap of , so that the person can sit themselves in the middle (seat of ) and not be next to anyone. Hence the minimum value of is .
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.