2013 AMC 12A Problems/Problem 21
Problem
Consider . Which of the following intervals contains ?
Solution 1
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:
Solution 2
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.
Solution 3
Define , and
We are looking for . We can 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 leaves only one answer , so we are done.
See Also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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