Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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− | == | + | == Problem == |
+ | Consider <math>A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots ))))</math>. Which of the following intervals contains <math>A</math>? | ||
− | Let <math>f(x) = \log(x + f(x-1))</math> and <math>f(2) = log(2)</math>, and from the problem description, <math>A = f(2013)</math> | + | <math> \textbf{(A)} \ (\log 2016, \log 2017) </math> |
+ | <math> \textbf{(B)} \ (\log 2017, \log 2018) </math> | ||
+ | <math> \textbf{(C)} \ (\log 2018, \log 2019) </math> | ||
+ | <math> \textbf{(D)} \ (\log 2019, \log 2020) </math> | ||
+ | <math> \textbf{(E)} \ (\log 2020, \log 2021) </math> | ||
+ | |||
+ | == Solutions == | ||
+ | === Solution 1 === | ||
+ | Let <math>f(x) = \log(x + f(x-1))</math> and <math>f(2) = \log(2)</math>, and from the problem description, <math>A = f(2013)</math> | ||
We can reason out an approximation, by ignoring the <math>f(x-1)</math>: | We can reason out an approximation, by ignoring the <math>f(x-1)</math>: | ||
− | <math> | + | <math>f_{0}(x) \approx \log x</math> |
And a better approximation, by plugging in our first approximation for <math>f(x-1)</math> in our original definition for <math>f(x)</math>: | And a better approximation, by plugging in our first approximation for <math>f(x-1)</math> in our original definition for <math>f(x)</math>: | ||
− | <math> | + | <math>f_{1}(x) \approx \log(x + \log(x-1))</math> |
And an even better approximation: | And an even better approximation: | ||
− | <math> | + | <math>f_{2}(x) \approx \log(x + \log(x-1 + \log(x-2)))</math> |
− | Continuing this pattern, obviously, will eventually terminate at our original definition of <math>f(x)</math>. | + | Continuing this pattern, obviously, will eventually terminate at <math>f_{x-1}(x)</math>, in other words our original definition of <math>f(x)</math>. |
− | However, at <math>x = 2013</math>, going further than | + | However, at <math>x = 2013</math>, going further than <math>f_{1}(x)</math> will not distinguish between our answer choices. <math>\log(2012 + \log(2011))</math> is nearly indistinguishable from <math>\log(2012)</math>. |
− | So we take | + | So we take <math>f_{1}(x)</math> and plug in. |
<math>f(2013) \approx \log(2013 + \log 2012)</math> | <math>f(2013) \approx \log(2013 + \log 2012)</math> | ||
− | Since <math>1000 < 2012 < 10000</math>, we know <math>3 < log(2012) < 4</math>. This gives us our answer range: | + | Since <math>1000 < 2012 < 10000</math>, we know <math>3 < \log(2012) < 4</math>. This gives us our answer range: |
<math>\log 2016 < \log(2013 + \log 2012) < \log(2017)</math> | <math>\log 2016 < \log(2013 + \log 2012) < \log(2017)</math> | ||
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<math>(\log 2016, \log 2017)</math> | <math>(\log 2016, \log 2017)</math> | ||
− | ==Solution 2== | + | === Solution 2 === |
+ | Suppose <math>A=\log(x)</math>. | ||
+ | Then <math>\log(2012+ \cdots)=x-2013</math>. | ||
+ | So if <math>x>2017</math>, then <math>\log(2012+\log(2011+\cdots))>4</math>. | ||
+ | So <math>2012+\log(2011+\cdots)>10000</math>. | ||
+ | Repeating, we then get <math>2011+\log(2010+\cdots)>10^{7988}</math>. | ||
+ | This is clearly absurd (the RHS continues to grow more than exponentially for each iteration). | ||
+ | So, <math>x</math> is not greater than <math>2017</math>. | ||
+ | So <math>A<\log(2017)</math>. | ||
+ | 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)), \text{ for } n > 2.</math> We are looking for <math>f(2013)</math>. First we show | ||
+ | |||
+ | '''Lemma.''' For any integer <math>n>2</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>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>. | ||
+ | |||
+ | === Solution 4 === | ||
+ | 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: | ||
+ | |||
+ | '''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>. | ||
+ | |||
+ | It follows that <math>\log(z+\log(x+\log(y))) < \log(x)+\log(y)+\log(z)</math>, and so on. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | === 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> | ||
+ | |||
+ | === Solution 6 (super quick) === | ||
+ | |||
+ | Let <math>f(x) = \log(x + \log((x-1) + \log((x-2) + \ldots + \log 2 \ldots )))</math>. From the answer choices, we see that <math>3 < f(2013) < 4</math>. Since <math>f(x)</math> grows very slowly, we can assume <math>3 < f(2012) < 4</math>. Therefore, <math>f(2013) = \log(2013 + f(2012)) \in (\log 2016, \log 2017) \implies \boxed{\textbf{(A)}}</math>. | ||
+ | |||
+ | ~rayfish | ||
+ | |||
+ | === Video Solution by Richard Rusczyk === | ||
+ | https://artofproblemsolving.com/videos/amc/2013amc12a/360 | ||
− | + | == See Also == | |
+ | {{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:51, 28 August 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
Solution 6 (super quick)
Let . From the answer choices, we see that . Since grows very slowly, we can assume . Therefore, .
~rayfish
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 |
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