2013 AMC 12A Problems/Problem 21
Consider . Which of the following intervals contains ?
Let and , and from the problem description,
We can reason out an approximation, by ignoring the :
And a better approximation, by plugging in our first approximation for in our original definition for :
And an even better approximation:
Continuing this pattern, obviously, will eventually terminate at , in other words our original definition of .
However, at , going further than will not distinguish between our answer choices. is nearly indistinguishable from .
So we take and plug in.
Since , we know . This gives us our answer range:
Suppose . Then . So if , then . So . Repeating, we then get . This is clearly absurd (the RHS continues to grow more than exponentially for each iteration). So, is not greater than . So . But this leaves only one answer, so we are done.
Define , and We are looking for . First we show
Lemma. For any integer , if then .
Proof. First note that . Let . Then , so . Suppose the claim is true for . Then . The Lemma is thus proved by induction.
Finally, note that so that the Lemma implies that . This means that , which leaves us with only one option .
Define , and We start with a simple observation:
Lemma. For , .
Proof. Since , we have , so .
It follows that , and so on.
It follows that .
Finally, we get , which leaves us with only option .
Solution 5 (nonrigorous + abusing answer choices.)
Intuitively, you can notice that , therefore (by the answer choices) . We can then say:
The only answer choice that is possible given this information is
Solution 6 (super quick)
Let . From the answer choices, we see that . Since grows very slowly, we can assume . Therefore, .
Video Solution by Richard Rusczyk
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