Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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So <math>A<\log(2017)</math>. | So <math>A<\log(2017)</math>. | ||
But this leaves only one answer, so we are done. | But this leaves only one answer, so we are done. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Define <math>f(2) = \log(2)</math>, and | ||
+ | |||
+ | <math>f(n) = \log(n+f(n-1)), \quad \text{ for } n > 2.</math> | ||
+ | |||
+ | We are looking for <math>f(2013)</math>. We can show | ||
+ | |||
+ | '''Lemma.''' For any integer <math>n>1</math>, if <math>n < 10^k - k</math> then <math>f(n) < k</math>. | ||
+ | |||
+ | '''Proof.''' First note that <math>f(2) < 1</math>. Let <math>n<10^k-k</math>. Then <math>n+k<10^k</math>, so <math>\log(n+k)< k</math>. Suppose the claim is true for <math>n-1</math>. Then <math>f(n) = \log(n+f(n-1)) < \log(n + k) < k</math>. The Lemma is thus proved by induction. | ||
+ | |||
+ | Finally, note that <math>2013 < 10^4 - 4</math> so that the Lemma implies that <math>f(2013) < 4</math>. | ||
+ | This leaves only one answer <math>\boxed{\textbf{(A) } (\log 2016, \log 2017)}</math>, so we are done. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:32, 17 December 2019
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 |
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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