AMC12 2012 A

AMC12 2012 A · Q24

AMC12 2012 A · Q24. It mainly tests Decimals, Patterns & sequences (misc).

Let $\{a_k\}_{k=1}^{2011}$ be the sequence of real numbers defined by $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$, and in general, \[a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}\] Rearranging the numbers in the sequence $\{a_k\}_{k=1}^{2011}$ in decreasing order produces a new sequence $\{b_k\}_{k=1}^{2011}$. What is the sum of all integers $k$, $1\le k \le 2011$, such that $a_k=b_k?$

设 $\{a_k\}_{k=1}^{2011}$ 为实数序列，定义为 $a_1=0.201,$ $a_2=(0.2011)^{a_1},$ $a_3=(0.20101)^{a_2},$ $a_4=(0.201011)^{a_3}$，一般地， \[a_k=\begin{cases}(0.\underbrace{20101\cdots 0101}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is odd,}\\(0.\underbrace{20101\cdots 01011}_{k+2\text{ digits}})^{a_{k-1}}\qquad\text{if }k\text{ is even.}\end{cases}\] 将序列 $\{a_k\}_{k=1}^{2011}$ 中的数按降序重排得到新序列 $\{b_k\}_{k=1}^{2011}$。所有满足 $a_k=b_k$ 的整数 $k$（$1\le k \le 2011$）之和是多少？

(A) 671 671

(B) 1006 1006

(D) 2011 2011

(E) 2012 2012

Answer

Correct choice: (C)

正确答案：（C）

Solution

First, we must understand two important functions: $f(x) = b^x$ for $0 < b < 1$(decreasing exponential function), and $g(x) = x^k$ for $k > 0$(increasing power function for positive $x$). $f(x)$ is used to establish inequalities when we change the exponent and keep the base constant. $g(x)$ is used to establish inequalities when we change the base and keep the exponent constant. We will now examine the first few terms. Comparing $a_1$ and $a_2$, $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$. Therefore, $0 < a_1 < a_2 < 1$. Comparing $a_2$ and $a_3$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$. Comparing $a_1$ and $a_3$, $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$. Therefore, $0 < a_1 < a_3 < a_2 < 1$. Comparing $a_3$ and $a_4$, $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$. Comparing $a_2$ and $a_4$, $0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$. Therefore, $0 < a_1 < a_3 < a_4 < a_2 < 1$. Continuing in this manner, it is easy to see a pattern(see Note 1). Therefore, the only $k$ when $a_k = b_k$ is when $2(k-1006) = 2011 - k$. Solving gives $\boxed{\textbf{(C)} 1341}$. Note 1: We claim that $0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1$. We can use induction to prove this statement. (not necessary for AMC): Base Case: We have already shown the base case above, where $0 < a_1 < a_2 < 1$. Inductive Step: Rearranging in decreasing order gives $1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b_{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} = a_1 > 0$.

首先，我们必须理解两个重要函数：当 $0 < b < 1$ 时，$f(x) = b^x$（递减的指数函数）；当 $k > 0$ 时，$g(x) = x^k$（对正 $x$ 递增的幂函数）。当改变指数而保持底数不变时，用 $f(x)$ 建立不等式；当改变底数而保持指数不变时，用 $g(x)$ 建立不等式。现在考察前几项。比较 $a_1$ 与 $a_2$，有 $0 < a_1 = (0.201)^1 < (0.201)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_1 < a_2 < 1$。因此，$0 < a_1 < a_2 < 1$。比较 $a_2$ 与 $a_3$，有 $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_3 < a_2 < 1$。比较 $a_1$ 与 $a_3$，有 $0 < a_1 = (0.201)^1 < (0.201)^{a_2} < (0.20101)^{a_2} = a_3 < 1 \Rightarrow 0 < a_1 < a_3 < 1$。因此，$0 < a_1 < a_3 < a_2 < 1$。比较 $a_3$ 与 $a_4$，有 $0 < a_3 = (0.20101)^{a_2} < (0.20101)^{a_3} < (0.201011)^{a_3} = a_4 < 1 \Rightarrow 0 < a_3 < a_4 < 1$。比较 $a_2$ 与 $a_4$，有 $0 < a_4 = (0.201011)^{a_3} < (0.201011)^{a_1} < (0.2011)^{a_1} = a_2 < 1 \Rightarrow 0 < a_4 < a_2 < 1$。因此，$0 < a_1 < a_3 < a_4 < a_2 < 1$。继续这样比较下去，很容易看出一个规律（见注 1）。因此，满足 $a_k = b_k$ 的 $k$ 只能在 $2(k-1006) = 2011 - k$ 时出现。解得 $\boxed{\textbf{(C)} 1341}$。注 1：我们断言 $0 < a_1 < a_3 < ... < a_{2011} < a_{2010} < ... < a_4 < a_2 < 1$。可用归纳法证明该结论（AMC 不要求）：基础情形：上面已证明 $0 < a_1 < a_2 < 1$。归纳步骤：按降序重排得到 $1 > b_1 = a_2 > b_2 = a_4 > ... > b_{1005} = a_{2010} > b_{1006} = a_{2011} > ... > b_{2010} = a_3 > b_{2011} = a_1 > 0$。

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