Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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Sequence: <math> 2, 3, 4, 6, 9, 14, 15, 16, 17 </math> | Sequence: <math> 2, 3, 4, 6, 9, 14, 15, 16, 17 </math> | ||
| − | Any number greater than <math> | + | Any number greater than <math> 10 </math> also cannot be a median of set <math> R </math>. |
There are then <math> \boxed{\textbf{(D)}\ 7} </math> possible medians of set <math> R </math>. | There are then <math> \boxed{\textbf{(D)}\ 7} </math> possible medians of set <math> R </math>. | ||
Revision as of 13:09, 1 October 2016
Problem
Let 
be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of ?
Solution
First, we find that the minimum value of the median of will be 
.
We then experiment with sequences of numbers to determine other possible medians.
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Median: 
Sequence: 
Any number greater than 
also cannot be a median of set .
There are then 
possible medians of set .
See Also
| 2012 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
