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

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==Solution 1==
 
==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>
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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>:

Revision as of 19:25, 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 (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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