2013 AMC 12A Problems/Problem 21

Revision as of 23:04, 13 February 2013 by Epicwisdom (talk | contribs) (Clarification and subscripts)


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:

$\log 2016 < \log(2013 + \log 2012) < \log(2017)$

$(\log 2016, \log 2017)$

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