Difference between revisions of "2009 AMC 12A Problems/Problem 24"

Revision as of 18:05, 11 February 2009

Problem

The tower function of twos is defined recursively as follows: $T(1) = 2$ and $T(n + 1) = 2^{T(n)}$ for $n\ge1$ . Let $A = (T(2009))^{T(2009)}$ and $B = (T(2009))^A$ . What is the largest integer $k$ such that

$\underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}$

is defined?

$\textbf{(A)}\ 2009\qquad \textbf{(B)}\ 2010\qquad \textbf{(C)}\ 2011\qquad \textbf{(D)}\ 2012\qquad \textbf{(E)}\ 2013$

Solution

We just look at the last three logarithms for the moment, and use the fact that $\log_2 T(k) = T(k - 1)$ . We wish to find:

\begin{align*}
& \log_2\left(\log_2\left(\log_2 \left({T(2009)^{\left({T(2009)}}^{T(2009)}\right)}\right)\right)\right) \\
&= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009)))) \\
&= \log_2(T(2009)\log_2(T(2009)T(2008))) \\
&= \log_2(T(2009)(T(2008) + T(2007))) (Error compiling LaTeX. Unknown error_msg)

Now we realize that $T(n - 1)$ is much smaller than $T(n)$ . So we approximate this, remembering we have rounded down, as:

$\log_2(T(2009)) = T(2008)$

We have used $3$ logarithms so far. Applying $2007$ more to the left of our expression, we get $T(1) = 2$ . Then we can apply the logarithm $2$ more times, until we get to $0$ . So our answer is approximately $3 + 2007 + 2 = 2012$ . But we rounded down, so that means that after $2012$ logarithms we get a number slightly greater than $0$ , so we can apply logarithms one more time. We can be sure it is small enough so that the logarithm can only be applied $1$ more time since $2012 + 1 = 2013$ is the largest answer choice. So the answer is $\mathbf{(E)}$ .

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "2009 AMC 12A Problems/Problem 24"

Revision as of 18:05, 11 February 2009

Problem

Solution

See also