Difference between revisions of "2021 AMC 12A Problems/Problem 18"
MRENTHUSIASM (talk | contribs) (→Solution 4 (Comprehensive, Similar to Solution 3)) |
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~yofro | ~yofro | ||
− | ==Solution 4 (Comprehensive, Similar to Solution 3)== | + | ==Solution 4 (Most Comprehensive, Similar to Solution 3)== |
We have the following important results: | We have the following important results: | ||
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Result <math>(3):</math> For all positive rational numbers <math>a,</math> we have <cmath>f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.</cmath> It follows that <math>f\left({\frac 1a}\right)=-f(a),</math> and result <math>(3)</math> is true. | Result <math>(3):</math> For all positive rational numbers <math>a,</math> we have <cmath>f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.</cmath> It follows that <math>f\left({\frac 1a}\right)=-f(a),</math> and result <math>(3)</math> is true. | ||
− | For all positive integers <math> | + | For all positive integers <math>x</math> and <math>y,</math> suppose <math>\prod_{k=1}^{n}{p_i}^{e_i}</math> and <math>\prod_{k=1}^{n}{q_i}^{d_i}</math> are their prime factorizations, respectively, we have |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | f\left(\frac | + | f\left(\frac xy\right)&=f\left(\frac {\prod_{k=1}^{n}{p_i}^{e_i}}{\prod_{k=1}^{n}{q_i}^{d_i}}\right) \\ |
+ | &=f\left(\prod_{k=1}^{n}\frac{{p_i}^{e_i}}{{q_i}^{d_i}}\right) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 00:26, 15 February 2021
- The following problem is from both the 2021 AMC 10A #18 and 2021 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
Let be a function defined on the set of positive rational numbers with the property that
for all positive rational numbers
and
. Furthermore, suppose that
also has the property that
for every prime number
. For which of the following numbers
is
?
Solution 1 (but where do you get ![$10=11+f(\frac{25}{11})$](//latex.artofproblemsolving.com/a/7/8/a785f1b14b898b8570bf90d64231c86c73d3ae0f.png)
Looking through the solutions we can see that can be expressed as
so using the prime numbers to piece together what we have we can get
, so
or
.
-Lemonie
- awesomediabrine
Solution 2
We know that . By transitive, we have
Subtracting
from both sides gives
Also
In
we have
.
In we have
.
In we have
.
In we have
.
In we have
.
Thus, our answer is
~JHawk0224 ~awesomediabrine
Solution 3 (Deeper)
Consider the rational , for
integers. We have
. So
. Let
be a prime. Notice that
. And
. So if
,
. We simply need this to be greater than what we have for
. Notice that for answer choices
and
, the numerator
has less prime factors than the denominator, and so they are less likely to work. We check
first, and it works, therefore the answer is
.
~yofro
Solution 4 (Most Comprehensive, Similar to Solution 3)
We have the following important results:
for all positive integers
for all positive rational numbers
Proofs
Result can be shown by induction.
Result For all positive rational numbers
we have
Therefore, we get
So, result
is true.
Result For all positive rational numbers
we have
It follows that
and result
is true.
For all positive integers and
suppose
and
are their prime factorizations, respectively, we have
~MRENTHUSIASM
Video Solution by Hawk Math
https://www.youtube.com/watch?v=dvlTA8Ncp58
Video Solution by Punxsutawney Phil
Video Solution by OmegaLearn (Using Functions and manipulations)
~ pi_is_3.14
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 10 Problems and Solutions |
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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.