Question

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 $a_k = b_k?$

$671$

$1006$

$1341$

$2011$

$2012$

Answer: C

Concepts:exponent inequality

Difficulty rating: 2460

Solution:

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 Problem 24

Answer: C

Difficulty rating: 2460

Solution:

Problem 24 in Other Years