2012 AMC 12A 第 24 题

Question

设

\{a_k\}_{k=1}^{2011}

为如下定义的实数序列：

a_1 = 0.201

，

a_2 = (0.2011)^{a_1}

，

a_3 = (0.20101)^{a_2}

，

a_4 = (0.201011)^{a_3}

。更一般地，

a_k = \begin{cases} \tiny \left(0.\underbrace{20101\ldots0101}_{k+2 \text{ 位数字}}\right)^{a_{k-1}}, & \tiny \text{若 } k \text{ 为奇数，} \\ \tiny \left(0.\underbrace{20101\ldots01011}_{k+2 \text{ 位数字}}\right)^{a_{k-1}}, & \tiny \text{若 } k \text{ 为偶数。} \end{cases}

将序列

\{a_k\}_{k=1}^{2011}

中的数按递减顺序重新排列，得到新序列

\{b_k\}_{k=1}^{2011}

。求所有满足

a_k = b_k

的整数

k

（其中

1 \le k \le 2011

）的和。

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},

and

a_4 = (0.201011)^{a_3},

and more generally

a_k = \begin{cases} \tiny \left(0.\underbrace{20101\ldots0101}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \text{if } k \text{ is odd,} \\ \tiny \left(0.\underbrace{20101\ldots01011}_{k+2 \text{ digits}}\right)^{a_{k-1}}, & \tiny \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 the integers

k,

1 \le k \le 2011,

such that

a_k = b_k?

671

1006

1341

2011

2012

答案：C

知识点：指数不等式

难度评级：2460

小提示：

每个底数都在

0

和

1

之间，所以幂运算会反转大小关系；比较相邻项

Each base is between

0

and

1,

so raising to a power reverses order; compare consecutive terms

大提示：

这些值交错排列：

1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010}

\gt a_{2011} \gt a_{2009}

\gt \cdots \gt a_1

；当某项的排名等于其下标时，它保持不动

The values interleave:

1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010}

\gt a_{2011} \gt a_{2009}

\gt \cdots \gt a_1;

a term is fixed when its rank equals its index

解答：

因为每个底数都严格介于

0

和

1

之间，函数

t \mapsto (\text{底数})^t

是递减的，而对

b \gt 0

，

t \mapsto t^b

是递增的。比较各项可得序列的大小顺序为

\begin{aligned} &1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \\ &\gt a_{2011} \gt a_{2009} \\ &\gt \cdots \gt a_1 \gt 0 \end{aligned}\text{。}

因此在递减排列中，偶数下标项先出现，然后是奇数下标项按反向出现。某项满足

a_k = b_k

当且仅当它的位置等于它的下标；对递减排列的奇数尾部，这要求

2(k - 1006) = 2011 - k

。

解得

3k = 4023

，所以

k = 1341

，这是唯一固定下标，因此所求和为

1341

。

因此，正确答案是 C。

Because each base lies strictly between

0

and

1,

the function

t \mapsto (\text{base})^t

is decreasing, while

t \mapsto t^b

is increasing for

b \gt 0.

Comparing terms shows the sequence orders as

\begin{aligned} &1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \\ &\gt a_{2011} \gt a_{2009} \\ &\gt \cdots \gt a_1 \gt 0. \end{aligned}

So in the decreasing arrangement, the even-indexed terms come first, then the odd-indexed terms in reverse. A term satisfies

a_k = b_k

exactly when its position equals its index, which for the descending odd tail requires

2(k - 1006) = 2011 - k.

Solving gives

3k = 4023,

so

k = 1341,

the unique fixed index, and the sum is

1341.

Thus, the correct answer is C.

2012 AMC 12A 第 24 题

答案：C

难度评级：2460

小提示：

大提示：

解答：

其他年份的第 24 题