Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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==Solution 2== | ==Solution 2== | ||
− | We know that <math>f( | + | We know that <math>f(p) = f(p \cdot 1) = f(p) + f(1)</math>. By Transitive, we have<cmath>f(p) &= f(p) + f(1).</cmath> |
− | <cmath> | + | Subtracting <math>f(p)</math> from both sides gives <math>0 = f(1).</math> |
− | f( | ||
− | |||
− | |||
− | f(1) | ||
− | |||
Also | Also | ||
<cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath> | <cmath>f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2</cmath> | ||
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Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math> | Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math> | ||
− | ~JHawk0224 | + | ~JHawk0224 -awesomediabrine |
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− | - awesomediabrine | ||
==Solution 3 (Deeper)== | ==Solution 3 (Deeper)== |
Revision as of 17:17, 14 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
\[f(p) &= f(p) + f(1).\] (Error compiling LaTeX. Unknown error_msg)
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
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.