Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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| + | https://youtu.be/yBSrLxv0LbY ~savannahsolver | ||
==See Also== | ==See Also== | ||
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Latest revision as of 08:36, 18 May 2022
Problem
Let 
be a set of nine distinct integers. Six of the elements are 
, 
, 
, 
, 
, and 
. 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 .
Therefore, the answer is 
Solution 2
Let the values of the missing integers be 
. We will find the bound of the possible medians.
The smallest possible median will happen when we order the set as 
. The median is .
The largest possible median will happen when we order the set as 
. The median is .
Therefore, the median must be between and inclusive, yielding possible medians, so the answer is 
.
~superagh
Video Solution
https://youtu.be/yBSrLxv0LbY ~savannahsolver
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. 
