Question

The tower function of twos is defined recursively as follows: $T(1) = 2$ and $T(n + 1) = 2^{T(n)}$ for $n \ge 1.$ Let $A = (T(2009))^{T(2009)}$ and $B = (T(2009))^A.$ What is the largest integer $k$ such that $\underbrace{\log_2 \log_2 \log_2 \ldots \log_2}_{k \text{ times}} B$ is defined?

$2009$

$2010$

$2011$

$2012$

$2013$

Solution:

Since $\log_2 T(n + 1) = T(n),$ each application of $\log_2$ strips one $2$ off the top of a tower of twos.

Reducing $B = (T(2009))^A$ with $A = (T(2009))^{T(2009)},$ one finds $\log_2 B = A\cdot T(2008),$ $\log_2^2 B = T(2009)T(2008) + T(2007),$ and in general the dominant term after $k + 3$ logs is $T(2008 - k).$

So after $2012$ applications of $\log_2$ the result is still positive, meaning a $2013$ th $\log_2$ is defined. A matching upper bound shows the result becomes negative after $2013$ applications, so a $2014$ th $\log_2$ is undefined. Hence the largest $k$ is $2013.$

Thus, the correct answer is E.

2009 AMC 12A Problem 24

Solution:

Problem 24 in Other Years