Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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We have the following important results: | We have the following important results: | ||
− | <math>(1) \ f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}a_k | + | <math>(1) \ f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}a_k</math> for all positive integers <math>k</math> |
<math>(2) \ f(1)=0</math> | <math>(2) \ f(1)=0</math> | ||
− | <math>(3) \ f\left({\frac 1a}\right)=-f(a) | + | <math>(3) \ f\left({\frac 1a}\right)=-f(a) for all positive rational numbers </math>a<math> |
<b>Proofs</b> | <b>Proofs</b> | ||
− | Result <math>(1)< | + | Result </math>(1)<math> can be shown by induction. |
− | Result <math>(2):< | + | Result </math>(2):<math> For all positive rational numbers </math>a,<math> we have <cmath>f(a)=f(a\cdot1)=f(a)+f(1).</cmath> Therefore, we get </math>f(1)=0.<math> So, result </math>(2)<math> is true. |
− | Result <math>(3):< | + | 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> So, </math>f\left({\frac 1a}\right)=-f(a),<math> and result </math>(3)$ is true. |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 00:54, 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 (Comprehensive, Similar to Solution 3)
We have the following important results:
for all positive integers
a$<b>Proofs</b>
Result$ (Error compiling LaTeX. Unknown error_msg)(1)$can be shown by induction.
Result$ (Error compiling LaTeX. Unknown error_msg)(2):a,
f(1)=0.
(2)$is true.
Result$ (Error compiling LaTeX. Unknown error_msg)(3):a,
f\left({\frac 1a}\right)=-f(a),
(3)$ is true.
~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.