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> |
− | + | <cmath>\log_2\log_2\log_2\left({T(2009)^{\left({T(2009)}^{T(2009)}\right)}}\right)</cmath> | |
− | + | <cmath>= \log_2(T(2009)\log_2(T(2009)\log_2 T(2009)))</cmath> | |
− | + | <cmath>= \log_2(T(2009)\log_2(T(2009)T(2008)))</cmath> | |
− | + | <cmath>= \log_2(T(2009)(T(2008) + T(2007)))</cmath> | |
− | + | </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 21:50, 2 May 2015
Problem
The tower function of twos is defined recursively as follows: and for . Let and . What is the largest integer such that
is defined?
Solution
We just look at the last three logarithms for the moment, and use the fact that . We wish to find:
Now we realize that is much smaller than . So we approximate this, remembering we have rounded down, as:
We have used logarithms so far. Applying more to the left of our expression, we get . Then we can apply the logarithm more times, until we get to . So our answer is approximately . But we rounded down, so that means that after logarithms we get a number slightly greater than , so we can apply logarithms one more time. We can be sure it is small enough so that the logarithm can only be applied more time since is the largest answer choice. So the answer is .
Alternative Solution
Let where there are 's. is defined iff iff . Note , so . Thus, we seek the largest such that . Now note that
so 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.