2013 AMC 12A Problems/Problem 21

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 $\boxed{\textbf{(A)}}$ .

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc12a/360

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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