Difference between revisions of "2021 AMC 12A Problems/Problem 18"
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− | {{duplicate|[[2021 AMC 10A Problems | + | {{duplicate|[[2021 AMC 10A Problems/Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems/Problem 18|2021 AMC 12A #18]]}} |
==Problem== | ==Problem== | ||
− | Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b) = f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. | + | Let <math>f</math> be a function defined on the set of positive rational numbers with the property that <math>f(a\cdot b)=f(a)+f(b)</math> for all positive rational numbers <math>a</math> and <math>b</math>. Suppose that <math>f</math> also has the property that <math>f(p)=p</math> for every prime number <math>p</math>. For which of the following numbers <math>x</math> is <math>f(x)<0</math>? |
− | <math>\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\ | + | <math>\textbf{(A) }\frac{17}{32} \qquad \textbf{(B) }\frac{11}{16} \qquad \textbf{(C) }\frac79 \qquad \textbf{(D) }\frac76\qquad \textbf{(E) }\frac{25}{11}</math> |
− | ==Solution 1== | + | ==Solution 1 (Intuitive)== |
− | + | From the answer choices, note that | |
+ | <cmath>\begin{align*} | ||
+ | f(25)&=f\left(\frac{25}{11}\cdot11\right) \\ | ||
+ | &=f\left(\frac{25}{11}\right)+f(11) \\ | ||
+ | &=f\left(\frac{25}{11}\right)+11. | ||
+ | \end{align*}</cmath> | ||
+ | On the other hand, we have | ||
+ | <cmath>\begin{align*} | ||
+ | f(25)&=f(5\cdot5) \\ | ||
+ | &=f(5)+f(5) \\ | ||
+ | &=5+5 \\ | ||
+ | &=10. | ||
+ | \end{align*}</cmath> | ||
+ | Equating the expressions for <math>f(25)</math> produces <cmath>f\left(\frac{25}{11}\right)+11=10,</cmath> from which <math>f\left(\frac{25}{11}\right)=-1.</math> Therefore, the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}.</math> | ||
− | + | <u><b>Remark</b></u> | |
− | + | Similarly, we can find the outputs of <math>f</math> at the inputs of the other answer choices: | |
− | + | <cmath>\begin{alignat*}{10} | |
− | + | &\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && 7 \\ | |
− | f( | + | &\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && 3 \\ |
− | + | &\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && 1 \\ | |
− | + | &\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && 2 | |
− | \end{ | + | \end{alignat*}</cmath> |
+ | Alternatively, refer to Solutions 2 and 4 for the full processes. | ||
+ | |||
+ | ~Lemonie ~awesomediabrine ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (Specific)== | ||
+ | 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> | ||
+ | Subtracting <math>f(p)</math> from both sides gives <math>0 = f(1).</math> | ||
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> | ||
Line 32: | Line 52: | ||
In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>. | In <math>\textbf{(E)}</math> we have <math>f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1</math>. | ||
− | Thus, our answer is <math>\boxed{\textbf{(E)} \frac{25}{11}}</math> | + | Thus, our answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}</math>. |
− | |||
− | ==Solution 3 ( | + | ~JHawk0224 ~awesomediabrine |
− | Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+ | + | |
+ | ==Solution 3 (Generalized)== | ||
+ | Consider the rational <math>\frac{a}{b}</math>, for <math>a,b</math> integers. We have <math>f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)</math>. So <math>f\left(\frac{a}{b}\right)=f(a)-f(b)</math>. Let <math>p</math> be a prime. Notice that <math>f(p^k)=kf(p)</math>. And <math>f(p)=p</math>. So if <math>a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}</math>, <math>f(a)=a_1p_1+a_2p_2+\cdots+a_kp_k</math>. We simply need this to be greater than what we have for <math>f(b)</math>. Notice that for answer choices <math>\textbf{(A)},\textbf{(B)},\textbf{(C)},</math> and <math>\textbf{(D)}</math>, the numerator has fewer prime factors than the denominator, and so they are less likely to work. We check <math>\textbf{(E)}</math> first, and it works, therefore the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}</math>. | ||
~yofro | ~yofro | ||
+ | |||
+ | ==Solution 4 (Generalized)== | ||
+ | We derive the following properties of <math>f:</math> | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>By induction, we have <cmath>f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)</cmath> for all positive rational numbers <math>a_k</math> and positive integers <math>n.</math> <p> | ||
+ | Since positive powers are just repeated multiplication of the base, it follows that <cmath>f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a)</cmath> for all positive rational numbers <math>a</math> and positive integers <math>n.</math></li><p> | ||
+ | <li>For all positive rational numbers <math>a,</math> we have <cmath>f(a)=f(a\cdot1)=f(a)+f(1),</cmath> from which <math>f(1)=0.</math></li><p> | ||
+ | <li>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> from which <math>f\left({\frac 1a}\right)=-f(a).</math></li><p> | ||
+ | </ol> | ||
+ | For all positive integers <math>x</math> and <math>y,</math> suppose <math>\prod_{k=1}^{m}p_k^{d_k}</math> and <math>\prod_{k=1}^{n}q_k^{e_k}</math> are their respective prime factorizations. We get | ||
+ | <cmath>\begin{align*} | ||
+ | f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) \\ | ||
+ | &=f(x)-f(y) && \hspace{10mm}\text{by Property 3} \\ | ||
+ | &=f\left(\prod_{k=1}^{m}p_k^{d_k}\right)-f\left(\prod_{k=1}^{n}q_k^{e_k}\right) \\ | ||
+ | &=\left[\sum_{k=1}^{m}f\left(p_k^{d_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{e_k}\right)\right] && \hspace{10mm}\text{by Property 1} \\ | ||
+ | &=\left[\sum_{k=1}^{m}d_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}e_k f\left(q_k\right)\right] && \hspace{10mm}\text{by Property 1} \\ | ||
+ | &=\left[\sum_{k=1}^{m}d_k p_k \right]-\left[\sum_{k=1}^{n}e_k q_k \right]. | ||
+ | \end{align*}</cmath> | ||
+ | We apply <math>f</math> to each fraction in the answer choices: | ||
+ | <cmath>\begin{alignat*}{10} | ||
+ | &\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && f\left(\frac{17^1}{2^5}\right) \quad && = \quad && [1(17)]-[5(2)] \quad && = \quad && 7 \\ | ||
+ | &\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && f\left(\frac{11^1}{2^4}\right) \quad && = \quad && [1(11)]-[4(2)] \quad && = \quad && 3 \\ | ||
+ | &\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && f\left(\frac{7^1}{3^2}\right) \quad && = \quad && [1(7)]-[2(3)] \quad && = \quad && 1 \\ | ||
+ | &\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && f\left(\frac{7^1}{2^1\cdot3^1}\right) \quad && = \quad && [1(7)]-[1(2)+1(3)] \quad && = \quad && 2 \\ | ||
+ | &\textbf{(E)} \qquad && f\left(\frac{25}{11}\right) \quad && = \quad && f\left(\frac{5^2}{11^1}\right) \quad && = \quad && [2(5)]-[1(11)] \quad && = \quad && {-}1 | ||
+ | \end{alignat*}</cmath> | ||
+ | Therefore, the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 5 (Quick, Dirty, and Frantic Last Hope)== | ||
+ | Note that answer choices <math>\textbf{(A)}</math> through <math>\textbf{(D)}</math> are <math>\frac{\text{prime}}{\text{composite}},</math> whereas <math>\textbf{(E)}</math> is <math>\frac{\text{composite}}{\text{prime}}.</math> Because the functional equation is related to primes, we hope that the uniqueness of answer choice <math>\boxed{\textbf{(E) }\frac{25}{11}}</math> is enough. | ||
+ | |||
+ | ~OliverA | ||
+ | |||
+ | ==Solution 6 (Rushed Generalization)== | ||
+ | If f(a <math>\cdot</math> b) = f(a) + f(b), and if f(p) = p, then f(p <math>\cdot</math> p) = 2p. You can do this multiple times (Ex: f(p^3) = 3p). You can quickly assume then, that f(p^n) = np. Thus the answer choices can then be rewritten as the product of a prime and another prime to the negative power. Answer choices A-C are straightforward. For D, you can rewrite <math>\frac{1}{6}</math> as <math>\frac{1}{2}</math> <math>\cdot</math> <math>\frac{1}{3}</math>. When you get to E, you get f(25) + f(<math>\frac{1}{11}</math>), which is 10 - 11, which is -1. So the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}</math> | ||
+ | |||
+ | ~Zeeshan12 | ||
==Video Solution by Hawk Math== | ==Video Solution by Hawk Math== | ||
https://www.youtube.com/watch?v=dvlTA8Ncp58 | https://www.youtube.com/watch?v=dvlTA8Ncp58 | ||
+ | |||
+ | ==Video Solution by North America Math Contest Go Go Go Through Induction== | ||
+ | https://www.youtube.com/watch?v=ffX0fTgJN0w&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=12 | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtu.be/8gGcj95rlWY | https://youtu.be/8gGcj95rlWY | ||
− | == Video Solution by OmegaLearn (Using Functions and | + | == Video Solution by OmegaLearn (Using Functions and Manipulations) == |
https://youtu.be/aGv99CLzguE | https://youtu.be/aGv99CLzguE | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/IUJ_A9KiLEE | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution (Quick and Easy)== | ||
+ | https://youtu.be/NbAu_STtcvA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2021|ab=A|num-b=17|num-a=19}} | ||
{{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}} | {{AMC12 box|year=2021|ab=A|num-b=17|num-a=19}} | ||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:30, 23 October 2022
- The following problem is from both the 2021 AMC 10A #18 and 2021 AMC 12A #18, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Intuitive)
- 3 Solution 2 (Specific)
- 4 Solution 3 (Generalized)
- 5 Solution 4 (Generalized)
- 6 Solution 5 (Quick, Dirty, and Frantic Last Hope)
- 7 Solution 6 (Rushed Generalization)
- 8 Video Solution by Hawk Math
- 9 Video Solution by North America Math Contest Go Go Go Through Induction
- 10 Video Solution by Punxsutawney Phil
- 11 Video Solution by OmegaLearn (Using Functions and Manipulations)
- 12 Video Solution by TheBeautyofMath
- 13 Video Solution (Quick and Easy)
- 14 See also
Problem
Let be a function defined on the set of positive rational numbers with the property that
for all positive rational numbers
and
. Suppose that
also has the property that
for every prime number
. For which of the following numbers
is
?
Solution 1 (Intuitive)
From the answer choices, note that
On the other hand, we have
Equating the expressions for
produces
from which
Therefore, the answer is
Remark
Similarly, we can find the outputs of at the inputs of the other answer choices:
Alternatively, refer to Solutions 2 and 4 for the full processes.
~Lemonie ~awesomediabrine ~MRENTHUSIASM
Solution 2 (Specific)
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 (Generalized)
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 fewer 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 (Generalized)
We derive the following properties of
- By induction, we have
for all positive rational numbers
and positive integers
Since positive powers are just repeated multiplication of the base, it follows that
for all positive rational numbers
and positive integers
- For all positive rational numbers
we have
from which
- For all positive rational numbers
we have
from which
For all positive integers and
suppose
and
are their respective prime factorizations. We get
We apply
to each fraction in the answer choices:
Therefore, the answer is
~MRENTHUSIASM
Solution 5 (Quick, Dirty, and Frantic Last Hope)
Note that answer choices through
are
whereas
is
Because the functional equation is related to primes, we hope that the uniqueness of answer choice
is enough.
~OliverA
Solution 6 (Rushed Generalization)
If f(a b) = f(a) + f(b), and if f(p) = p, then f(p
p) = 2p. You can do this multiple times (Ex: f(p^3) = 3p). You can quickly assume then, that f(p^n) = np. Thus the answer choices can then be rewritten as the product of a prime and another prime to the negative power. Answer choices A-C are straightforward. For D, you can rewrite
as
. When you get to E, you get f(25) + f(
), which is 10 - 11, which is -1. So the answer is
~Zeeshan12
Video Solution by Hawk Math
https://www.youtube.com/watch?v=dvlTA8Ncp58
Video Solution by North America Math Contest Go Go Go Through Induction
https://www.youtube.com/watch?v=ffX0fTgJN0w&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=12
Video Solution by Punxsutawney Phil
Video Solution by OmegaLearn (Using Functions and Manipulations)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
Video Solution (Quick and Easy)
~Education, the Study of Everything
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.