Difference between revisions of "2013 AMC 12A Problems/Problem 21"
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The only answer choice that is possible given this information is <math>\boxed{\textbf{(A)}}</math> | The only answer choice that is possible given this information is <math>\boxed{\textbf{(A)}}</math> | ||
+ | |||
+ | === Solution 6 (super quick) === | ||
+ | |||
+ | Let <math>f(x) = \log(x + \log((x-1) + \log((x-2) + \ldots + \log 2 \ldots )))</math>. From the answer choices, we see that <math>3 < f(2013) < 4</math>. Since <math>f(x)</math> grows very slowly, we can assume <math>3 < f(2012) < 4</math>. Therefore, <math>f(2013) = \log(2013 + f(2012)) \in (\log 2016, \log 2017) \implies \boxed{\textbf{(A)}}</math>. | ||
+ | |||
+ | ~rayfish | ||
+ | |||
+ | === Solution 7 (Not rigorous but it works) === | ||
+ | Start small from <math>log(3+log2)</math>. Because <math>log2</math> is close to <math>0</math>, the expression is close to <math>log 3</math>. This continues for the 1 digit numbers. However, when we get to <math>log(11+log10)</math>, <math>log10=1</math> so the expression equals <math>log12</math>. We can round down how much each new log contributes to the expression as an approximate answer. There are 90 2 digit numbers, 900 3 digit numbers, and 1014 4 digit numbers that increase the expression. <math>1000< 90 \cdot 1 + 900 \cdot 2 + 1014 \cdot 3< 10000</math>, so it increases between 3 and 4. The answer is <math>\boxed{\textbf{(A)}}</math>. | ||
+ | |||
+ | ~dragnin | ||
=== Video Solution by Richard Rusczyk === | === Video Solution by Richard Rusczyk === |
Latest revision as of 15:48, 9 June 2023
Contents
Problem
Consider . Which of the following intervals contains
?
Solutions
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
.
Solution 5 (nonrigorous + abusing answer choices.)
Intuitively, you can notice that , therefore (by the answer choices)
. We can then say:
The only answer choice that is possible given this information is
Solution 6 (super quick)
Let . From the answer choices, we see that
. Since
grows very slowly, we can assume
. Therefore,
.
~rayfish
Solution 7 (Not rigorous but it works)
Start small from . Because
is close to
, the expression is close to
. This continues for the 1 digit numbers. However, when we get to
,
so the expression equals
. We can round down how much each new log contributes to the expression as an approximate answer. There are 90 2 digit numbers, 900 3 digit numbers, and 1014 4 digit numbers that increase the expression.
, so it increases between 3 and 4. The answer is
.
~dragnin
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2013amc12a/360
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.