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 20:25, 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 |
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