AMC8 2012

AMC8 2012 · Q22

AMC8 2012 · Q22. It mainly tests Casework, Logic puzzles.

Let $R$ be a set of nine distinct integers. Six of the elements of the set are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of $R$?

令$R$是一个包含九个不同整数的集合。其中六个元素是2、3、4、6、9和14。$R$的中位数的可能值有多少个？

(A) 4 4

(B) 5 5

(D) 7 7

(E) 8 8

Answer

Correct choice: (D)

正确答案：（D）

Solution

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.

令缺失的整数值为$x, y, z$。我们将找出可能中位数的界限。最小可能中位数发生在集合按$\{x, y, z, 2, 3, 4, 6, 9, 14\}$排序时。中位数是$3$。最大可能中位数发生在集合按$\{2, 3, 4, 6, 9, 14, x, y, z\}$排序时。中位数是$9$。因此，中位数必须在$3$和$9$之间（包含两端），从而有$\boxed{\textbf{(D)}\ 7}$个可能中位数。

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