Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 12A Problems/Problem 21"
(→‎= Solution 9 (i just lost my dawg))
 
(25 intermediate revisions by 15 users not shown)
Line 1: Line 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>
+
== 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>?
 +
 
 +
<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>f(x) \approx \log x</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>f(x) \approx \log(x + \log(x-1))</math>
+
<math>f_{1}(x) \approx \log(x + \log(x-1))</math>
  
 
And an even better approximation:
 
And an even better approximation:
  
−
<math>f(x) \approx \log(x + \log(x-1 + \log(x-2)))</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 our second approximation will not distinguish between our answer choices.
+
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 our second approximation and plug in.
+
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>
  
 
<math>(\log 2016, \log 2017)</math>
 
<math>(\log 2016, \log 2017)</math>
 +
 +
=== 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
 +
 +
=== Solution 7 (Not rigorous but it works) ===
 +
Start small from <math>log(3+log2)</math>. Because <math>log2</math> is close to <math>0</math>, the expression is close to <math>log 3</math>. This continues for the 1 digit numbers. However, when we get to <math>log(11+log10)</math>, <math>log10=1</math> so the expression equals <math>log12</math>. We can round down how much each new log contributes to the expression as an approximate answer. There are 90 2 digit numbers, 900 3 digit numbers, and 1014 4 digit numbers that increase the expression. <math>1000< 90 \cdot 1 + 900 \cdot 2 + 1014 \cdot 3< 10000</math>, so it increases between 3 and 4. The answer is <math>\boxed{\textbf{(A)}}</math>.
 +
 +
~dragnin
 +
 +
=== Solution 8 (Assumption) ===
 +
Consider <math>\log (2013+\log(2012))</math> and disregard every other log. We do this because we assume that the logs from <math>2011</math> onward cannot contribute more than <math>7987</math> because of the nature of logs. We have <math>\log(2016)<\log (2013+\log(2012))<\log(2017)</math> so the answer is <math>\boxed{A}.</math>
 +
 +
 +
=== Solution 9 (i just lost my dawg) ==
 +
 +
It seems that <math>\log (2012+x)</math> is going to be less than four, since otherwise, it would imply that <math>x>7987</math>, and going down, that seems very impossible.
 +
 +
Then, the answer is just <math>\boxed{\textbf{A} (\log{2016}, \log{2017})}</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}}
 +
{{MAA Notice}}

Latest revision as of 17:38, 21 September 2024

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

Solution 7 (Not rigorous but it works)

Start small from $log(3+log2)$. Because $log2$ is close to $0$, the expression is close to $log 3$. This continues for the 1 digit numbers. However, when we get to $log(11+log10)$, $log10=1$ so the expression equals $log12$. We can round down how much each new log contributes to the expression as an approximate answer. There are 90 2 digit numbers, 900 3 digit numbers, and 1014 4 digit numbers that increase the expression. $1000< 90 \cdot 1 + 900 \cdot 2 + 1014 \cdot 3< 10000$, so it increases between 3 and 4. The answer is $\boxed{\textbf{(A)}}$.

~dragnin

Solution 8 (Assumption)

Consider $\log (2013+\log(2012))$ and disregard every other log. We do this because we assume that the logs from $2011$ onward cannot contribute more than $7987$ because of the nature of logs. We have $\log(2016)<\log (2013+\log(2012))<\log(2017)$ so the answer is $\boxed{A}.$


= Solution 9 (i just lost my dawg)

It seems that $\log (2012+x)$ is going to be less than four, since otherwise, it would imply that $x>7987$, and going down, that seems very impossible.

Then, the answer is just $\boxed{\textbf{A} (\log{2016}, \log{2017})}$

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_21&oldid=227735"