2012 AMC 8 Problems/Problem 22

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 1

First, we find that the minimum value of the median of $R$ will be $3$.

We then experiment with sequences of numbers to determine other possible medians.

Median: $3$

Sequence: $-2, -1, 0, 2, 3, 4, 6, 9, 14$

Median: $4$

Sequence: $-1, 0, 2, 3, 4, 6, 9, 10, 14$

Median: $5$

Sequence: $0, 2, 3, 4, 5, 6, 9, 10, 14$

Median: $6$

Sequence: $0, 2, 3, 4, 6, 9, 10, 14, 15$

Median: $7$

Sequence: $2, 3, 4, 6, 7, 8, 9, 10, 14$

Median: $8$

Sequence: $2, 3, 4, 6, 8, 9, 10, 14, 15$

Median: $9$

Sequence: $2, 3, 4, 6, 9, 14, 15, 16, 17$

Any number greater than $9$ also cannot be a median of set $R$.

Therefore, the answer is $7\implies \textbf{(D)}.$

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 $7$ possible medians, so the answer is $\textbf{(D)}$.


~superagh

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
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

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