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

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== Solution ==
 
== Solution ==
 
We just look at the last three logarithms for the moment, and use the fact that <math>\log_2 T(k) = T(k - 1)</math>. We wish to find:
 
We just look at the last three logarithms for the moment, and use the fact that <math>\log_2 T(k) = T(k - 1)</math>. We wish to find:
−
<center><cmath>
+
<center>
−
\begin{align*}
+
<cmath>\log_2\log_2\log_2\left({T(2009)^{\left({T(2009)}^{T(2009)}\right)}}\right)</cmath>
−
& \log_2\left(\log_2\left(\log_2 \left({T(2009)^{\left({T(2009)}}^{T(2009)}\right)}\right)\right)\right) \\
+
<cmath>= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009)))</cmath>
−
&= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009))) \\
+
<cmath>= \log_2(T(2009)\log_2(T(2009)T(2008)))</cmath>
−
&= \log_2(T(2009)\log_2(T(2009)T(2008))) \\
+
<cmath>= \log_2(T(2009)(T(2008) + T(2007)))</cmath>
−
&= \log_2(T(2009)(T(2008) + T(2007)))</cmath></center>
+
</center>
 +
 
  
 
Now we realize that <math>T(n - 1)</math> is much smaller than <math>T(n)</math>. So we approximate this, remembering we have rounded down, as:
 
Now we realize that <math>T(n - 1)</math> is much smaller than <math>T(n)</math>. So we approximate this, remembering we have rounded down, as:

Revision as of 20:50, 2 May 2015

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:

\[\log_2\log_2\log_2\left({T(2009)^{\left({T(2009)}^{T(2009)}\right)}}\right)\] \[= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009)))\] \[= \log_2(T(2009)\log_2(T(2009)T(2008)))\] \[= \log_2(T(2009)(T(2008) + T(2007)))\]


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 $\mathbf{(E)}$.

Alternative Solution

Let $L_k(x)=\log_2\log_2...\log_2(x)$ where there are $k$ $\log_2$'s. $L_k(B)$ is defined iff $L_{k-1}(B) > 0$ iff $L_{k-2}(B) > 1$. Note $\log_2 T(k) = T(k - 1)$, so $L_{k-2}(T(k-2))=1$. Thus, we seek the largest $k$ such that $B > T(k-2)$. Now note that

$T(2009)^{T(2009)^{T(2009)}} > 2^{2^{T(2009)}} = T(2011)$

so $k=2013$ satisfies the inequality. Since it is the largest choice, it is the answer.

See also

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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