2009 AMC 12A Problems/Problem 24
Contents
Problem
The tower function of twos is defined recursively as follows: and for . Let and . What is the largest integer for which is defined?
Solution (not a proof, but good enough for AMCs)
Testing the first two (or three) positive integers instead of 2009, seems to always be 4 more. Put E and go on to tackle #25 :) ~ dolphin7
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 .
EDIT: Why use answer choices? @above solution should explain at the end WHY they can be sure.
Alternative Solution
Let where there are 's. is defined iff (if and only if) 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.
Clarification
(Note: This for the people who need clear, concise reasoning in the form of mostly words, instead of a compact proof written in set theory symbols and the like. I figured if I had trouble understanding the above solutions, others would too.)
We begin by contemplating what actually is. Calculating the first few values of , we see that it quickly becomes absolutely, massively, well, MASSIVE. So actually calculating , is sadly, not an option (this is a final five problem, after all). Next we consider if it is possible to express differently. We have:
(I didn't actually try plug it into a computer program; the point is, it's not that easy)
So, the only thing left to do is jump right into the problem. We start by applying the s one at a time, and seeing where it goes. We have:
(Uhhhhhh....)
But wait! We can use the rule . Then the expression becomes
(because )
Now we try ing it again:
.
This time, we can make use of the addition rule . The expression becomes
Pulling out the exponent again and further simplifying, we have
Now we it a third time... and we're stuck 😐. There is no pretty formula for taking the of two things added together. So we rack our brains, sigh in frustration... and give up.
JUST KIDDING! (Don't give up)
Whenever you get stuck, you want to "zoom out" and re-gain perspective of what it is you're actually trying to find. The problem asks us to find the number of times we can keep taking the of this expression, until we no longer can (once you reach a negative value, you can't take the log of it anymore; i.e. s are only defined for positive inputs). So, for the context of this problem, what we want to know is:
"How many times can I take the of and get a positive value?"
Let's think about what actually means. We want a number , such that when is raised to the power of , we get the expression in the . So, when we think about
What actually matters? Well, even if you have a big number, say, , what happens when you raise to the power of that number? We would end up getting another one of these:
Calculating... ERROR
(lol 🤣)
But in all seriousness, whenever we consider tower functions, the next tower number is always much, much bigger. So, back to the problem:
is huge... but when you put it next to , it looks like an ant. Even more so when it's next to . So... what does that mean? Well, if we could get rid of the , we could at least go another step. So... let's get rid of it! (At this point I can almost see the audience: *surprised pikachu face)
(Just kidding, but yes, the following part of the explanation does incorporate some calculus-related ideas)
If we were to consider just
we would get
But what if we wanted to get
as our answer? What would the argument have to be? (In this context, "argument" refers to the value in the parentheses). Well, we can work backwards:
So the argument would have to be at least
But when we compare
to
Clearly
So,
equals some value between
and
Let's call that small excess value (lowercase epsilon, literally "infinitely small but not 0"). We want to keep taking the of the expression, and is going to be pitifully small compared to or . So, within the context of this problem,
Now we take the again:
Look familiar? It should, because this is almost exactly the same scenario as the one I just made a big fuss about (Take that, you no-know-calculus noobs! On an off note, it's completely fine if you don't know calculus; AMC problems are supposed to solvable by students who don't know calculus.) So, applying the same idea as earlier we have
From here on out, it's calm waters. To recap,
Taking the more times(for a total of ), we get
Taking the log again, we get
(Now we're at times)
Once more and we have
.
( times)
But wait, there's more! (IYKYK lol) Remember how we ignored the ? Well, It's time to bring it back. Every time we took the log, we would get + because we defined to be some very, very small (but finite) number. Let's go back a few steps:
should really be
and
should really be
Since is still a positive number, we can take the of it one last time before it becomes negative. So, after a long and very tiring journey (for both you, the reader, and me, the writer) we have (the number of times we took the ) .
~TheAsian
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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