AMC10 2000 A

AMC10 2000 A · Q23

AMC10 2000 A · Q23. It mainly tests Averages (mean), Arithmetic misc.

When the mean, median, and mode of the list $10, 2, 5, 2, 4, 2, x$ are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real value of $x$?

当列表 $10, 2, 5, 2, 4, 2, x$ 的均值、中位数和众数按升序排列时，它们形成一个非恒等的等差数列。所有可能的实数 $x$ 的和是多少？

(A) 3 3

(B) 6 6

(D) 17 17

(E) 20 20

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): If $x$ were less than or equal to $2$, then $2$ would be both the median and the mode of the list. Thus $x>2$. Consider the two cases $2<x<4$, and $x\ge 4$. Case 1: If $2<x<4$, then $2$ is the mode, $x$ is the median, and $\frac{25+x}{7}$ is the mean, which must equal $2-(x-2)$, $\frac{x+2}{2}$, or $x+(x-2)$, depending on the size of the mean relative to $2$ and $x$. These give $x=\frac{3}{8}$, $x=\frac{36}{5}$, and $x=3$, of which $x=3$ is the only value between $2$ and $4$. Case 2: If $x\ge 4$, then $4$ is the median, $2$ is the mode, and $\frac{25+x}{7}$ is the mean, which must be $0,3$, or $6$. Thus $x=-25$, $-4$, or $17$, of which $17$ is the only one of these values greater than or equal to $4$. Thus the $x$-value sum to $3+17=20$.

答案（E）：如果 $x\le 2$，那么 $2$ 将同时是该列表的中位数和众数。因此 $x>2$。考虑两种情况：$2<x<4$，以及 $x\ge 4$。情况 1：若 $2<x<4$，则 $2$ 是众数，$x$ 是中位数，且 $\frac{25+x}{7}$ 是平均数。根据平均数相对于 $2$ 和 $x$ 的大小，它必须等于 $2-(x-2)$、$\frac{x+2}{2}$ 或 $x+(x-2)$。解得 $x=\frac{3}{8}$、$x=\frac{36}{5}$、$x=3$，其中只有 $x=3$ 落在 $2$ 与 $4$ 之间。情况 2：若 $x\ge 4$，则 $4$ 是中位数，$2$ 是众数，且 $\frac{25+x}{7}$ 是平均数，它必须为 $0$、$3$ 或 $6$。因此 $x=-25$、$-4$ 或 $17$，其中只有 $17$ 大于或等于 $4$。因此 $x$ 的取值之和为 $3+17=20$。

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