Difference between revisions of "2013 AMC 12A Problems/Problem 21"

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(Solution 6 (super quick))
 
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==Problem==
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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>?
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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>?
  
 
<math> \textbf{(A)} \ (\log 2016, \log 2017) </math>
 
<math> \textbf{(A)} \ (\log 2016, \log 2017) </math>
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<math> \textbf{(E)} \ (\log 2020, \log 2021) </math>
 
<math> \textbf{(E)} \ (\log 2020, \log 2021) </math>
  
 
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== Solutions ==
==Solution 1==
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=== Solution 1 ===
 
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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>
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>:
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<math>(\log 2016, \log 2017)</math>
 
<math>(\log 2016, \log 2017)</math>
  
==Solution 2==
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=== Solution 2 ===
 
 
 
Suppose <math>A=\log(x)</math>.  
 
Suppose <math>A=\log(x)</math>.  
 
Then <math>\log(2012+ \cdots)=x-2013</math>.  
 
Then <math>\log(2012+ \cdots)=x-2013</math>.  
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But this leaves only one answer, so we are done.
 
But this leaves only one answer, so we are done.
  
==See Also==
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=== Solution 3 ===
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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 are looking for <math>f(2013)</math>. First we show
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'''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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'''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.
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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>.
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=== 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:
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'''Lemma.''' For <math>x,y>2</math>, <math>\log(x+\log(y)) < \log(x)+\log(y)=\log(xy)</math>.
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'''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>.
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Then <math>f(2011) = \log(2011+f(2010)) < \log(10047) < 5</math>.
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It follows that <math>f(2012) = \log(2012+f(2011)) < \log(2017) < 4</math>.
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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>.
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=== Solution 5 (nonrigorous + abusing answer choices.) ===
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 +
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:
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<cmath>x=\log(2013+\log(2012+\cdots(\log(2))\cdots))</cmath>
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<cmath>\log(2013)<x<\log(2013+\log(2021))</cmath>
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<cmath>\log(2013)<x<\log(2013+4)</cmath>
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<cmath>\log(2013)<x<\log(2017)</cmath>
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The only answer choice that is possible given this information is <math>\boxed{\textbf{(A)}}</math>
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=== Solution 6 (super quick) ===
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 +
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
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 +
=== Video Solution by Richard Rusczyk ===
 +
https://artofproblemsolving.com/videos/amc/2013amc12a/360
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== 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}}

Latest revision as of 14:51, 28 August 2021

Problem

Consider $A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots ))))$. Which of the following intervals contains $A$?

$\textbf{(A)} \ (\log 2016, \log 2017)$ $\textbf{(B)} \ (\log 2017, \log 2018)$ $\textbf{(C)} \ (\log 2018, \log 2019)$ $\textbf{(D)} \ (\log 2019, \log 2020)$ $\textbf{(E)} \ (\log 2020, \log 2021)$

Solutions

Solution 1

Let $f(x) = \log(x + f(x-1))$ and $f(2) = \log(2)$, and from the problem description, $A = f(2013)$

We can reason out an approximation, by ignoring the $f(x-1)$:

$f_{0}(x) \approx \log x$

And a better approximation, by plugging in our first approximation for $f(x-1)$ in our original definition for $f(x)$:

$f_{1}(x) \approx \log(x + \log(x-1))$

And an even better approximation:

$f_{2}(x) \approx \log(x + \log(x-1 + \log(x-2)))$

Continuing this pattern, obviously, will eventually terminate at $f_{x-1}(x)$, in other words our original definition of $f(x)$.

However, at $x = 2013$, going further than $f_{1}(x)$ will not distinguish between our answer choices. $\log(2012 + \log(2011))$ is nearly indistinguishable from $\log(2012)$.

So we take $f_{1}(x)$ and plug in.

$f(2013) \approx \log(2013 + \log 2012)$

Since $1000 < 2012 < 10000$, we know $3 < \log(2012) < 4$. This gives us our answer range:

$\log 2016 < \log(2013 + \log 2012) < \log(2017)$

$(\log 2016, \log 2017)$

Solution 2

Suppose $A=\log(x)$. Then $\log(2012+ \cdots)=x-2013$. So if $x>2017$, then $\log(2012+\log(2011+\cdots))>4$. So $2012+\log(2011+\cdots)>10000$. Repeating, we then get $2011+\log(2010+\cdots)>10^{7988}$. This is clearly absurd (the RHS continues to grow more than exponentially for each iteration). So, $x$ is not greater than $2017$. So $A<\log(2017)$. But this leaves only one answer, so we are done.

Solution 3

Define $f(2) = \log(2)$, and $f(n) = \log(n+f(n-1)), \text{ for } n > 2.$ We are looking for $f(2013)$. First we show

Lemma. For any integer $n>2$, if $n < 10^k-k$ then $f(n) < k$.

Proof. First note that $f(2) < 1$. Let $n<10^k-k$. Then $n+k<10^k$, so $\log(n+k)< k$. Suppose the claim is true for $n-1$. Then $f(n) = \log(n+f(n-1)) < \log(n + k) < k$. The Lemma is thus proved by induction.

Finally, note that $2012 < 10^4 - 4$ so that the Lemma implies that $f(2012) < 4$. This means that $f(2013) = \log(2013+f(2012)) < \log(2017)$, which leaves us with only one option $\boxed{\textbf{(A) } (\log 2016, \log 2017)}$.

Solution 4

Define $f(2) = \log(2)$, and $f(n) = \log(n+f(n-1)), \text{ for } n > 2.$ We start with a simple observation:

Lemma. For $x,y>2$, $\log(x+\log(y)) < \log(x)+\log(y)=\log(xy)$.

Proof. Since $x,y>2$, we have $xy-x-y = (x-1)(y-1) - 1 > 0$, so $xy - x-\log(y) > xy - x - y > 0$.

It follows that $\log(z+\log(x+\log(y))) < \log(x)+\log(y)+\log(z)$, and so on.

Thus $f(2010) < \log 2 + \log 3 + \cdots + \log 2010 < \log 2010 + \cdots + \log 2010 <  2009\cdot 4 = 8036$.

Then $f(2011) = \log(2011+f(2010)) < \log(10047) < 5$.

It follows that $f(2012) = \log(2012+f(2011)) < \log(2017) < 4$.

Finally, we get $f(2013) = \log(2013 + f(2012)) < \log(2017)$, which leaves us with only option $\boxed{\textbf{(A)}}$.

Solution 5 (nonrigorous + abusing answer choices.)

Intuitively, you can notice that $\log(2012+\log(2011+\cdots(\log(2)\cdots))<\log(2013+\log(2012+\cdots(\log(2))\cdots))$, therefore (by the answer choices) $\log(2012+\log(2011+\cdots(\log(2)\cdots))<\log(2021)$. We can then say:

\[x=\log(2013+\log(2012+\cdots(\log(2))\cdots))\] \[\log(2013)<x<\log(2013+\log(2021))\] \[\log(2013)<x<\log(2013+4)\] \[\log(2013)<x<\log(2017)\]

The only answer choice that is possible given this information is $\boxed{\textbf{(A)}}$

Solution 6 (super quick)

Let $f(x) = \log(x + \log((x-1) + \log((x-2) + \ldots + \log 2 \ldots )))$. From the answer choices, we see that $3 < f(2013) < 4$. Since $f(x)$ grows very slowly, we can assume $3 < f(2012) < 4$. Therefore, $f(2013) = \log(2013 + f(2012)) \in (\log 2016, \log 2017) \implies \boxed{\textbf{(A)}}$.

~rayfish

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc12a/360

See Also

2013 AMC 12A (ProblemsAnswer KeyResources)
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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