Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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+ | ==Problem== | ||
+ | Consider <math> A = \log (2013 + \log (2012 + \log (2011 + \log (\cdots + \log (3 + \log 2) \cdots )))) </math>. Which of the following intervals contains <math> A </math>? | ||
+ | |||
+ | <math> \textbf{(A)} \ (\log 2016, \log 2017) </math> | ||
+ | <math> \textbf{(B)} \ (\log 2017, \log 2018) </math> | ||
+ | <math> \textbf{(C)} \ (\log 2018, \log 2019) </math> | ||
+ | <math> \textbf{(D)} \ (\log 2019, \log 2020) </math> | ||
+ | <math> \textbf{(E)} \ (\log 2020, \log 2021) </math> | ||
+ | |||
+ | |||
==Solution== | ==Solution== | ||
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==See Also== | ==See Also== | ||
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{{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2013|ab=A|num-b=20|num-a=22}} |
Revision as of 17:53, 22 February 2013
Problem
Consider . Which of the following intervals contains ?
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:
See Also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |