AMC12 2012 A

AMC12 2012 A · Q8

AMC12 2012 A · Q8. It mainly tests Manipulating equations, Averages (mean).

An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?

对数字 1、2、3、4 和 5 进行迭代平均，方法如下：将这五个数按某种顺序排列。先求前两个数的平均数，再将该平均数与第三个数求平均，再将所得结果与第四个数求平均，最后将所得结果与第五个数求平均。用此过程可能得到的最大值与最小值之差是多少？

(A) \frac{31}{16} \frac{31}{16}

(B) 2 2

(D) 3 3

(E) \frac{65}{16} \frac{65}{16}

Answer

Correct choice: (C)

正确答案：（C）

Solution

The iterative average of any 5 integers $a,b,c,d,e$ is defined as: \[\frac{\frac{\frac{\frac{a+b} 2+c} 2+d} 2+e} 2=\frac{a+b+2c+4d+8e}{16}\] Plugging in $1,2,3,4,5$ for $a,b,c,d,e$, we see that in order to maximize the fraction, $a=1,b=2,c=3,d=4,e=5$, and in order to minimize the fraction, $a=5,b=4,c=3,d=2,e=1$. After plugging in these values and finding the positive difference of the two fractions, we arrive with $\frac{34}{16} \Rightarrow \frac{17}{8}$, which is our answer of $\boxed{\textbf{(C)}}$

任意 5 个整数 $a,b,c,d,e$ 的迭代平均定义为： \[\frac{\frac{\frac{\frac{a+b} 2+c} 2+d} 2+e} 2=\frac{a+b+2c+4d+8e}{16}\] 将 $1,2,3,4,5$ 分别代入 $a,b,c,d,e$，可见为了使该分数最大，取 $a=1,b=2,c=3,d=4,e=5$，为了使该分数最小，取 $a=5,b=4,c=3,d=2,e=1$。代入并求两者的正差，得到 $\frac{34}{16} \Rightarrow \frac{17}{8}$，因此答案为 $\boxed{\textbf{(C)}}$。

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