Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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Finally, we get <math>f(2013) = \log(2013 + f(2012)) < \log(2017)</math>, which leaves us with only option <math>\boxed{\textbf{(A)}}</math>. | Finally, we get <math>f(2013) = \log(2013 + f(2012)) < \log(2017)</math>, which leaves us with only option <math>\boxed{\textbf{(A)}}</math>. | ||
+ | |||
+ | === Solution 5 (nonrigorous + abusing answer choices.) === | ||
+ | |||
+ | Intuitively, you can notice that <math>\log(2012+\log(2011+\cdots(\log(2)\cdots))<\log(2013+\log(2012+\cdots(\log(2))\cdots))</math>, therefore (by the answer choices) <math>\log(2012+\log(2011+\cdots(\log(2)\cdots))<\log(2021)</math>. We can then say: | ||
+ | |||
+ | <cmath>x=\log(2013+\log(2012+\cdots(\log(2))\cdots))</cmath> | ||
+ | <cmath>\log(2013)<x<\log(2013+\log(2021))</cmath> | ||
+ | <cmath>\log(2013)<x<\log(2013+4)</cmath> | ||
+ | <cmath>\log(2013)<x<\log(2017)</cmath> | ||
+ | |||
+ | The only answer choice that is possible given this information is <math>\boxed{\textbf{(A)}}</math> | ||
=== Video Solution by Richard Rusczyk === | === Video Solution by Richard Rusczyk === |
Latest revision as of 12:16, 27 January 2021
Contents
Problem
Consider . Which of the following intervals contains ?
Solutions
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 . 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 .
Solution 4
Define , and We start with a simple observation:
Lemma. For , .
Proof. Since , we have , so .
It follows that , and so on.
Thus .
Then .
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
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc12a/360
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.