Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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==Solution 2== | ==Solution 2== | ||
For those who don't like bashing problems out, use this method. | For those who don't like bashing problems out, use this method. | ||
| − | 2,3,4,6,9,14,x,y,z | + | 2, 3, 4, 6,9, 14, x, y, z |
if we put x, y, z in front of 2, 3 can be the median | if we put x, y, z in front of 2, 3 can be the median | ||
Revision as of 00:34, 3 November 2017
Contents
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 1
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: 
Median: 
Sequence: 
Any number greater than also cannot be a median of set .
There are then 
possible medians of set .
Solution 2
For those who don't like bashing problems out, use this method. 2, 3, 4, 6,9, 14, x, y, z
if we put x, y, z in front of 2, 3 can be the median
if we put x, y, z after 14, 9 is the median
therefore, 3,4,6,9,x,y,z can all be the medians
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. 
