Question

一组八个整数的平均数、中位数、唯一众数和极差都等于 $8$ 。这组数中可能出现的最大整数是多少？

The mean, median, unique mode, and range of a collection of eight integers are all equal to $8.$ The largest integer that can be an element of this collection is

$11$

$12$

$13$

$14$

$15$

答案：D

知识点：平均数极差极端原理

难度评级：1660

解答：

集合 $6, 6, 6, 8, 8, 8, 8, 14$ 的平均数、中位数、唯一众数和极差都等于 $8$ ，所以 $14$ 可以达到。

假设最大值为 $15$ 。极差 $8$ 迫使最小值为 $7$ ，中位数 $8$ 又固定中间两个值为 $8, 8$ 。于是 $7 + 8 + 8 + 15 = 38$ ，剩下四个值的和为 $64 - 38 = 26$ ，平均为 $6.5$ 。至少有一个会小于 $7$ ，与最小值矛盾。因此 $15$ 不可能。 $8$ ，

因此，正确答案是 D。

The collection $6, 6, 6, 8, 8, 8, 8, 14$ has mean, median, unique mode, and range all equal to $8,$ so $14$ is attainable.

Suppose the largest were $15.$ The range $8$ forces the smallest to be $7.$ Because $8$ is the mode, it occurs in the sorted list; together with median $8,$ this forces the two middle values to be $8, 8.$ Then $7 + 8 + 8 + 15 = 38,$ so the remaining four values sum to $64 - 38 = 26,$ averaging $6.5.$ At least one would be below $7,$ contradicting the minimum. So $15$ is impossible.

Thus, the correct answer is D.

2002 AMC 12A 第 15 题

答案：D

难度评级：1660

解答：

其他年份的第 15 题