Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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<math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8 </math> | <math> \textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8 </math> | ||
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==Solution 2== | ==Solution 2== | ||
Revision as of 14:50, 25 September 2025
Contents
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 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.
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
