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：平均数中位数（数据）众数极端原理

难度评级：1660

解答：

八个整数的和是 $8\cdot 8=64.$ 数列 $6,6,6,8,8,8,8,14$ 的平均数、中位数、唯一众数和极差都等于 $8,$ 所以最大值 $14$ 可以达到。

若最大值至少为 $16,$ 极差条件会使最小值至少为 $8.$ 平均数为 $8$ 就会迫使八个整数全都等于 $8,$ 与极差条件矛盾。

若最大值为 $15,$ 极差 $8$ 会使最小值为 $7,$ 所以八个整数都至少为 $7.$ 其余七个数的和是 $64-15=49=7\cdot 7,$ 迫使它们全都等于 $7.$ 但这样中位数和众数会是 $7,$ 而不是 $8,$ 矛盾。

所以正确答案是 D。

The sum is $8\cdot 8=64.$ 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.

If the largest were at least $16,$ the range condition would make the smallest at least $8.$ A mean of $8$ would then force all eight integers to equal $8,$ contradicting the range.

If the largest were $15,$ the range $8$ forces the smallest to be $7,$ so all eight integers are at least $7.$ The other seven then sum to $64-15=49=7\cdot 7,$ forcing every one of them to equal $7.$ But then the median and mode would be $7,$ not $8,$ a contradiction.

Thus, the correct answer is D.

2002 AMC 10A 第 21 题

答案：D

难度评级：1660

解答：

其他年份的第 21 题