2022 AMC 12B 第 7 题

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

小提示：

众数大于中位数，所以它只能出现在两个最大的数中。

The mode exceeds the median, so it can only appear among the two largest numbers

大提示：

设中位数为

m

；为了有唯一众数，两个最大数都必须等于

m+2

，且总和为

5(m-2)

。

Let the median be

m;

for a unique mode the two largest must both equal

m+2,

and the sum is

5(m-2)

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

小提示：

大提示：

解答：

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