Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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+ | ==Solution== | ||
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Let <math>f(x) = \log(x + f(x-1))</math> and <math>f(2) = log(2)</math>, and from the problem description, <math>A = f(2013)</math> | Let <math>f(x) = \log(x + f(x-1))</math> and <math>f(2) = log(2)</math>, and from the problem description, <math>A = f(2013)</math> | ||
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<math>(\log 2016, \log 2017)</math> | <math>(\log 2016, \log 2017)</math> | ||
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+ | ==Solution 2== | ||
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+ | We see that <math>10^A = 2013 + \cdots</math>, and that the quantity within "<math>\cdots</math>" is between <math>\log(1000)</math> and <math>\log(10000)</math>, i.e. <math>(3,4)</math>, therefore <math>10^A \in (2016,2017)</math>, or <math>A\in (\log(2016),\log(2017))</math>. |
Revision as of 12:15, 12 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 our original definition of .
However, at , going further than our second approximation will not distinguish between our answer choices.
So we take our second approximation and plug in.
Since , we know . This gives us our answer range:
Solution 2
We see that , and that the quantity within "" is between and , i.e. , therefore , or .