2013 AMC 12A Problems/Problem 21: Difference between revisions

Revision as of 00:32, 17 December 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.

Solution 3

Define $f(2) = \log(2)$ , and

$f(n) = \log(n+f(n-1)), \quad \text{ for } n > 2.$

We are looking for $f(2013)$ . We can show

Lemma. For any integer $n>1$ , 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 $2013 < 10^4 - 4$ so that the Lemma implies that $f(2013) < 4$ . This leaves only one answer $\boxed{\textbf{(A) } (\log 2016, \log 2017)}$ , so we are done.

@@ Line 50: / Line 50: @@
 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)), \quad \text{ for } n > 2.</math>
+We are looking for <math>f(2013)</math>. We can show
+'''Lemma.''' For any integer <math>n>1</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>2013 < 10^4 - 4</math> so that the Lemma implies that <math>f(2013) < 4</math>.
+This leaves only one answer <math>\boxed{\textbf{(A) } (\log 2016, \log 2017)}</math>, so we are done.
 ==See Also==
 {{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}}
 {{MAA Notice}}

2013 AMC 12A (Problems • Answer Key • Resources)
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