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2012 AMC 8 Problems/Problem 22: Difference between revisions

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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>
==Solution 1==
First, we find that the minimum value of the median of <math> R </math> will be <math> 3 </math>.
We then experiment with sequences of numbers to determine other possible medians.
Median: <math> 3 </math>
Sequence: <math> -2, -1, 0, 2, 3, 4, 6, 9, 14 </math>
Median: <math> 4 </math>
Sequence: <math> -1, 0, 2, 3, 4, 6, 9, 10, 14 </math>
Median: <math> 5 </math>
Sequence: <math> 0, 2, 3, 4, 5, 6, 9, 10, 14 </math>
Median: <math> 6 </math>
Sequence: <math> 0, 2, 3, 4, 6, 9, 10, 14, 15 </math>
Median: <math> 7 </math>
Sequence: <math> 2, 3, 4, 6, 7, 8, 9, 10, 14 </math>
Median: <math> 8 </math>
Sequence: <math> 2, 3, 4, 6, 8, 9, 10, 14, 15 </math>
Median: <math> 9 </math>
Sequence: <math> 2, 3, 4, 6, 9, 14, 15, 16, 17 </math>
Any number greater than <math> 9 </math> also cannot be a median of set <math> R </math>.
Therefore, the answer is <math>\boxed{\textbf{(D)}\ 7}</math>.


==Solution 2==
==Solution 2==
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The largest possible median will happen when we order the set as <math>\{2, 3, 4, 6, 9, 14, x, y, z\}</math>. The median is <math>9</math>.
The largest possible median will happen when we order the set as <math>\{2, 3, 4, 6, 9, 14, x, y, z\}</math>. The median is <math>9</math>.


Therefore, the median must be between <math>3</math> and <math>9</math> inclusive, yielding <math>7</math> possible medians, so the answer is <math>\textbf{(D)}</math>.  
Therefore, the median must be between <math>3</math> and <math>9</math> inclusive, yielding <math>\boxed{\textbf{(D)}\ 7}</math> possible medians.
 
 


~superagh
~superagh
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https://youtu.be/yBSrLxv0LbY ~savannahsolver
https://youtu.be/yBSrLxv0LbY ~savannahsolver


==See Also==
==
{{AMC8 box|year=2012|num-b=21|num-a=23}}
{{MAA Notice}}

Latest revision as of 15:13, 25 September 2025

Problem

Let $R$ 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 $R$?

$\textbf{(A)}\hspace{.05in}4\qquad\textbf{(B)}\hspace{.05in}5\qquad\textbf{(C)}\hspace{.05in}6\qquad\textbf{(D)}\hspace{.05in}7\qquad\textbf{(E)}\hspace{.05in}8$

Solution 2

Let the values of the missing integers be $x, y, z$. We will find the bound of the possible medians.

The smallest possible median will happen when we order the set as $\{x, y, z, 2, 3, 4, 6, 9, 14\}$. The median is $3$.

The largest possible median will happen when we order the set as $\{2, 3, 4, 6, 9, 14, x, y, z\}$. The median is $9$.

Therefore, the median must be between $3$ and $9$ inclusive, yielding $\boxed{\textbf{(D)}\ 7}$ possible medians.

~superagh

Video Solution

https://youtu.be/yBSrLxv0LbY ~savannahsolver

==

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