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 $1000 2017$, 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 $A 2.$ We are looking for $f(2013)$. First we show
Lemma. For any integer $n>2$, if $n  2.$ We start with a simple observation:
Lemma. For $x,y>2$, $\log(x+\log(y)) 2$, 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

Video Solution by Richard Rusczyk
[https://artofproblemsolving.com/videos/amc/2013amc12a/360](https://artofproblemsolving.comhttps://artofproblemsolving.com/videos/amc/2013amc12a/360)