Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
| − | 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> |
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>: | ||
Revision as of 19:25, 4 February 2019
Contents
Problem
Consider 
. Which of the following intervals contains 
?
Solution 1
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:
Solution 2
Suppose 
.
Then 
.
So if 
, then 
.
So 
.
Repeating, we then get 
.
This is clearly absurd (the RHS continues to grow more than exponentially for each iteration).
So, 
is not greater than 
.
So 
.
But this leaves only one answer, so we are done.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
