Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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==Solution 3== | ==Solution 3== | ||
− | Define <math>f(2) = \log(2)</math>, and | + | Define <math>f(2) = \log(2)</math>, and <math>f(n) = \log(n+f(n-1)), \text{ for } n > 2.</math> We are looking for <math>f(2013)</math>. First we show |
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− | <math>f(n) = \log(n+f(n-1)), | ||
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− | We are looking for <math>f(2013)</math>. | ||
'''Lemma.''' For any integer <math>n>2</math>, if <math>n < 10^k-k</math> then <math>f(n) < k</math>. | '''Lemma.''' For any integer <math>n>2</math>, if <math>n < 10^k-k</math> then <math>f(n) < k</math>. | ||
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Finally, note that <math>2012 < 10^4 - 4</math> so that the Lemma implies that <math>f(2012) < 4</math>. This means that <math>f(2013) = \log(2013+f(2012)) < \log(2017)</math>, which leaves us with only one option <math>\boxed{\textbf{(A) } (\log 2016, \log 2017)}</math>. | Finally, note that <math>2012 < 10^4 - 4</math> so that the Lemma implies that <math>f(2012) < 4</math>. This means that <math>f(2013) = \log(2013+f(2012)) < \log(2017)</math>, which leaves us with only one option <math>\boxed{\textbf{(A) } (\log 2016, \log 2017)}</math>. | ||
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+ | ==Solution 4== | ||
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+ | |||
+ | Define <math>f(2) = \log(2)</math>, and <math>f(n) = \log(n+f(n-1)), \text{ for } n > 2.</math> We start with a simple observation: | ||
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+ | '''Lemma.''' For <math>x,y>2</math>, <math>\log(x+\log(y)) < \log(x)+\log(y)=\log(xy)</math>. | ||
+ | |||
+ | '''Proof.''' Since <math>x,y>2</math>, we have <math>xy-x-y = (x-1)(y-1) - 1 > 0</math>, so <math>xy - x-\log(y) > xy - x - y > 0</math>. | ||
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+ | It follows that <math>\log(z+\log(x+\log(y))) < \log(x)+\log(y)+\log(z)</math>, and so on. | ||
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+ | Thus <math>f(2010) < \log 2 + \log 3 + \cdots + \log 2010 < \log 2010 + \cdots + \log 2010 < 2009\cdot 4 = 8036</math>. | ||
+ | |||
+ | Then <math>f(2011) = \log(2011+f(2010)) < \log(10047) < 5</math>. | ||
+ | |||
+ | It follows that <math>f(2012) = \log(2012+f(2011)) < \log(2017) < 4</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>. | ||
+ | |||
==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 22:41, 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 . 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 .
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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