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

(Solution 2 is equivalent to the original solution, and assumes the answer a priori)
(Clarification and subscripts)
Line 5: Line 5:
 
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>:
  
−
<math>f(x) \approx \log x</math>
+
<math>f_{0}(x) \approx \log x</math>
  
 
And a better approximation, by plugging in our first approximation for <math>f(x-1)</math> in our original definition for <math>f(x)</math>:
 
And a better approximation, by plugging in our first approximation for <math>f(x-1)</math> in our original definition for <math>f(x)</math>:
  
−
<math>f(x) \approx \log(x + \log(x-1))</math>
+
<math>f_{1}(x) \approx \log(x + \log(x-1))</math>
  
 
And an even better approximation:
 
And an even better approximation:
  
−
<math>f(x) \approx \log(x + \log(x-1 + \log(x-2)))</math>
+
<math>f_{2}(x) \approx \log(x + \log(x-1 + \log(x-2)))</math>
  
−
Continuing this pattern, obviously, will eventually terminate at our original definition of <math>f(x)</math>.
+
Continuing this pattern, obviously, will eventually terminate at <math>f_{x-1}(x)</math>, in other words our original definition of <math>f(x)</math>.
  
−
However, at <math>x = 2013</math>, going further than our second approximation will not distinguish between our answer choices.
+
However, at <math>x = 2013</math>, going further than <math>f_{1}(x)</math> will not distinguish between our answer choices. <math>\log(2012 + \log(2011))</math> is nearly indistinguishable from <math>\log(2012)</math>.
  
−
So we take our second approximation and plug in.
+
So we take <math>f_{1}(x)</math> and plug in.
  
 
<math>f(2013) \approx \log(2013 + \log 2012)</math>
 
<math>f(2013) \approx \log(2013 + \log 2012)</math>

Revision as of 22:04, 13 February 2013

Solution

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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