Difference between revisions of "2012 AMC 8 Problems/Problem 22"

Latest revision as of 16:13, 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

==

@@ Line 3: / Line 3: @@
 <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==
@@ Line 55: / Line 18: @@
 https://youtu.be/yBSrLxv0LbY ~savannahsolver
-==See Also==
+==
-{{AMC8 box|year=2012|num-b=21|num-a=23}}
-{{MAA Notice}}

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Difference between revisions of "2012 AMC 8 Problems/Problem 22"

Latest revision as of 16:13, 25 September 2025

Problem

Solution 2

Video Solution