Difference between revisions of "2016 AMC 10A Problems/Problem 13"
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<cmath>B,A,C,D,E\rightarrow A,C,B,E,D</cmath> | <cmath>B,A,C,D,E\rightarrow A,C,B,E,D</cmath> | ||
| − | So <math>A</math>, Ada, was originally in seat <math>\boxed{\textbf{B}\text{ 2}}</math>. | + | So <math>A</math>, Ada, was originally in seat <math>\boxed{\textbf{(B)}\text{ 2}}</math>. |
==See Also== | ==See Also== | ||
Revision as of 20:30, 4 February 2016
Contents
Problem
Five friends sat in a movie theater in a row containing 
seats, numbered 
to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?
Solution 1
Bash: we see that the following configuration works.
Bea - Ada - Ceci - Dee - Edie
After moving, it becomes
Ada - Ceci - Bea - Edie - Dee.
Thus, Ada was in seat 
.
Solution 2
Process of elimination of possible configurations.
Let's say that Ada=
, Bea=
, Ceci=
, Dee=
, and Edie=
.
Since moved more to the right than did left, this implies that was in a LEFT end seat originally:
![\[B,-,C\rightarrow -,C,B\]](http://latex.artofproblemsolving.com/5/e/b/5ebe425cf4c6c024fadf751169e5b6e87aadaf15.png)
This is affirmed because 
, which there is no new seats uncovered. So 
are restricted to the same 
seats.
Thus, it must be 
, and more specifically:
![\[B,A,C,D,E\rightarrow A,C,B,E,D\]](http://latex.artofproblemsolving.com/e/f/2/ef29a8a3442e37c77cd4075b33573e46338138c7.png)
So , Ada, was originally in seat 
.
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 AMC 10 Problems and Solutions | ||
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
