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

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==Problem==
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== Problem ==
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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>?
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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 ==
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==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>
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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>
 
  
 
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>f(2013) \approx \log(2013 + \log 2012)</math>
 
<math>f(2013) \approx \log(2013 + \log 2012)</math>
  
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Since <math>1000 < 2012 < 10000</math>, we know <math>3 < log(2012) < 4</math>. This gives us our answer range:
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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>
  
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==Solution 2==
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=== Solution 2 ===
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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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So if <math>x>2017</math>, then <math>\log(2012+log(2011+\cdots))>4</math>.  
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So if <math>x>2017</math>, then <math>\log(2012+\log(2011+\cdots))>4</math>.  
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So <math>2012+log(2011+\cdots)>10000</math>.  
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So <math>2012+\log(2011+\cdots)>10000</math>.  
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Repeating, we then get <math>2011+log(2010+\cdots)>10^{7988}</math>.  
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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).
 
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>x</math> is not greater than <math>2017</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.
  
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==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:
 +
 
 +
'''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>.
 +
 
 +
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.) ===
 +
 
 +
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>
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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>
 +
 
 +
The only answer choice that is possible given this information is <math>\boxed{\textbf{(A)}}</math>
 +
 
 +
=== 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}}
 
{{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:16, 27 January 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 $\boxed{\textbf{(A)}}$

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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