2012 AMC 12A Problem 24

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

Each base is between

0

and

1,

so raising to a power reverses order; compare consecutive terms

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

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.

Problem 24 in other years