Question

Camila 写下五个正整数。这些整数的唯一众数比它们的中位数大 $2$ ，而中位数比它们的算术平均数大 $2$ 。众数的最小可能值是多少？

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

$5$

$7$

$9$

$11$

$13$

答案：D

知识点：众数中位数（数据）平均数最优化

难度评级：1380

解答：

将五个数按递增排列，设中位数为 $m$ 。众数为 $m + 2 \gt m$ 所以它只能出现在两个最大数中；要成为唯一众数，这两个数必须都等于 $m + 2$ 。

平均数为 $m - 2$ 所以五个数之和为 $5(m-2)$ 除去两个 $m + 2$ 和中位数 $m$ 两个最小数之和为 $5(m-2) - m - 2(m+2)$ $= 2m - 14$ 。

两个最小数是不同的正整数，所以 $2m - 14 \ge 1 + 2 = 3$ 得 $m \ge 9$ 当 $m = 9$ 时， $1, 3, 9, 11, 11$ 满足条件，因此最小众数为 $m + 2 = 11$ 。

所以正确答案是 D。

List the numbers in increasing order with median $m.$ The mode is $m + 2 \gt m,$ so it can only occur among the two largest entries; for it to be the unique mode, both of them must equal $m + 2.$

The mean is $m - 2,$ so the total is $5(m-2).$ With the two largest equal to $m + 2$ and the median $m,$ the two smallest sum to $5(m-2) - m - 2(m+2)$ $= 2m - 14.$

The two smallest are distinct positive integers, so $2m - 14 \ge 1 + 2 = 3,$ giving $m \ge 9.$ With $m = 9$ the list $1, 3, 9, 11, 11$ works, so the least mode is $m + 2 = 11.$

Thus, the correct answer is D.

2022 AMC 12B 第 7 题

答案：D

难度评级：1380

解答：

其他年份的第 7 题