AMC10 2022 B

AMC10 2022 B · Q10

AMC10 2022 B · Q10. It mainly tests Averages (mean), Arithmetic misc.

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?

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 $2$，中位数比它们的算术平均数大 $2$。众数的最小可能值为多少？

(A) 5 5

(B) 7 7

(D) 11 11

(E) 13 13

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $M$ be the median. It follows that the two largest integers are both $M+2.$ Let $a$ and $b$ be the two smallest integers such that $a<b.$ The sorted list is \[a,b,M,M+2,M+2.\] Since the median is $2$ greater than their arithmetic mean, we have $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M,$ or \[a+b+14=2M.\] Note that $a+b$ must be even. We minimize this sum so that the arithmetic mean, the median, and the unique mode are minimized. Let $a=1$ and $b=3,$ from which $M=9$ and $M+2=\boxed{\textbf{(D)}\ 11}.$

设 $M$ 为中位数。那么两个最大整数均为 $M+2$。设两个最小整数为 $a<b$，排序列表为 \[a,b,M,M+2,M+2.\] 由于中位数比算术平均数大 $2$，有 $\frac{a+b+M+(M+2)+(M+2)}{5}+2=M$，即 \[a+b+14=2M.\] 注意 $a+b$ 必须为偶数。为使算术平均数、中位数和唯一众数最小，令 $a=1$，$b=3$，得 $M=9$，$M+2=\boxed{\textbf{(D)}\ 11}$。

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