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)
m
Line 33: Line 33:
 
<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>
Line 43: Line 43:
 
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>.  
−
So if <math>x>2017</math>, then <math>\log(2012+log(2011+\cdots))>4</math>.  
+
So if <math>x>2017</math>, then <math>\log(2012+\log(2011+\cdots))>4</math>.  
−
So <math>2012+log(2011+\cdots)>10000</math>.  
+
So <math>2012+\log(2011+\cdots)>10000</math>.  
−
Repeating, we then get <math>2011+log(2010+\cdots)>10^{7988}</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).
 
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>.  

Revision as of 20:25, 4 February 2019

Problem

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

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


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.

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=101246"