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

Epicwisdom (talk | contribs) (Solution 2 is equivalent to the original solution, and assumes the answer a priori) |
Epicwisdom (talk | contribs) (Clarification and subscripts) |
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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> | + | <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> | + | <math>f_{1}(x) \approx \log(x + \log(x-1))</math> |

And an even better approximation: | And an even better approximation: | ||

− | <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 | + | 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 | + | 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 23:04, 13 February 2013

## Solution

Let and , and from the problem description,

We can reason out an approximation, by ignoring the :

And a better approximation, by plugging in our first approximation for in our original definition for :

And an even better approximation:

Continuing this pattern, obviously, will eventually terminate at , in other words our original definition of .

However, at , going further than will not distinguish between our answer choices. is nearly indistinguishable from .

So we take and plug in.

Since , we know . This gives us our answer range: