2013 AMC 12A Problems/Problem 21
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.
Solution 3
Define , and
We are looking for
. First we show
Lemma. For any integer , if
then
.
Proof. First note that . Let
. Then
, so
. Suppose the claim is true for
. Then
. The Lemma is thus proved by induction.
Finally, note that so that the Lemma implies that
. This means that
, which leaves us with only one option
.
Solution 4
Define , and
We start with a simple observation:
Lemma. For ,
.
Proof. Since , we have
, so
.
It follows that , and so on.
Thus .
Then .
It follows that .
Finally, we get , which leaves us with only option
.
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=-3d31c-7xm8
~dolphin7
See Also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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All AMC 12 Problems and Solutions |
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