Difference between revisions of "2021 AMC 12A Problems/Problem 18"

Latest revision as of 07:19, 23 October 2024

The following problem is from both the 2021 AMC 10A #18 and 2021 AMC 12A #18, so both problems redirect to this page.

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 Solution 7 (Generalized)
9 Video Solution by Hawk Math
10 Video Solution by North America Math Contest Go Go Go Through Induction
11 Video Solution by Punxsutawney Phil
12 Video Solution by OmegaLearn (Using Functions and Manipulations)
13 Video Solution by TheBeautyofMath
14 Video Solution (Quick and Easy)
15 See also

Problem

Let $f$ be a function defined on the set of positive rational numbers with the property that $f(a\cdot b)=f(a)+f(b)$ for all positive rational numbers $a$ and $b$ . Suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$ . For which of the following numbers $x$ is $f(x)<0$ ?

$\textbf{(A) }\frac{17}{32} \qquad \textbf{(B) }\frac{11}{16} \qquad \textbf{(C) }\frac79 \qquad \textbf{(D) }\frac76\qquad \textbf{(E) }\frac{25}{11}$

Solution 1 (Intuitive)

From the answer choices, note that $\begin{align*} f(25)&=f\left(\frac{25}{11}\cdot11\right) \\ &=f\left(\frac{25}{11}\right)+f(11) \\ &=f\left(\frac{25}{11}\right)+11. \end{align*}$ On the other hand, we have $\begin{align*} f(25)&=f(5\cdot5) \\ &=f(5)+f(5) \\ &=5+5 \\ &=10. \end{align*}$ Equating the expressions for $f(25)$ produces $f\left(\frac{25}{11}\right)+11=10,$ from which $f\left(\frac{25}{11}\right)=-1.$ Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$

Remark

Similarly, we can find the outputs of $f$ at the inputs of the other answer choices: $\begin{alignat*}{10} &\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && 7 \\ &\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && 3 \\ &\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && 1 \\ &\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && 2 \end{alignat*}$ Alternatively, refer to Solutions 2 and 4 for the full processes.

~Lemonie ~awesomediabrine ~MRENTHUSIASM

Solution 2 (Specific)

We know that $f(p) = f(p \cdot 1) = f(p) + f(1)$ . By transitive, we have $f(p) = f(p) + f(1).$ Subtracting $f(p)$ from both sides gives $0 = f(1).$ Also $f(2)+f\left(\frac{1}{2}\right)=f(1)=0 \implies 2+f\left(\frac{1}{2}\right)=0 \implies f\left(\frac{1}{2}\right) = -2$ $f(3)+f\left(\frac{1}{3}\right)=f(1)=0 \implies 3+f\left(\frac{1}{3}\right)=0 \implies f\left(\frac{1}{3}\right) = -3$ $f(11)+f\left(\frac{1}{11}\right)=f(1)=0 \implies 11+f\left(\frac{1}{11}\right)=0 \implies f\left(\frac{1}{11}\right) = -11$ In $\textbf{(A)}$ we have $f\left(\frac{17}{32}\right)=17+5f\left(\frac{1}{2}\right)=17-5(2)=7$ .

In $\textbf{(B)}$ we have $f\left(\frac{11}{16}\right)=11+4f\left(\frac{1}{2}\right)=11-4(2)=3$ .

In $\textbf{(C)}$ we have $f\left(\frac{7}{9}\right)=7+2f\left(\frac{1}{3}\right)=7-2(3)=1$ .

In $\textbf{(D)}$ we have $f\left(\frac{7}{6}\right)=7+f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)=7-2-3=2$ .

In $\textbf{(E)}$ we have $f\left(\frac{25}{11}\right)=10+f\left(\frac{1}{11}\right)=10-11=-1$ .

Thus, our answer is $\boxed{\textbf{(E) }\frac{25}{11}}$ .

~JHawk0224 ~awesomediabrine

Solution 3 (Generalized)

Consider the rational $\frac{a}{b}$ , for $a,b$ integers. We have $f(a)=f\left(\frac{a}{b}\cdot b\right)=f\left(\frac{a}{b}\right)+f(b)$ . So $f\left(\frac{a}{b}\right)=f(a)-f(b)$ . Let $p$ be a prime. Notice that $f(p^k)=kf(p)$ . And $f(p)=p$ . So if $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ , $f(a)=a_1p_1+a_2p_2+\cdots+a_kp_k$ . We simply need this to be greater than what we have for $f(b)$ . Notice that for answer choices $\textbf{(A)},\textbf{(B)},\textbf{(C)},$ and $\textbf{(D)}$ , the numerator has fewer prime factors than the denominator, and so they are less likely to work. We check $\textbf{(E)}$ first, and it works, therefore the answer is $\boxed{\textbf{(E) }\frac{25}{11}}$ .

~yofro

Solution 4 (Generalized)

We derive the following properties of $f:$

By induction, we have $f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)$ for all positive rational numbers $a_k$ and positive integers $n.$
Since positive powers are just repeated multiplication of the base, it follows that $f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a)$ for all positive rational numbers $a$ and positive integers $n.$

For all positive rational numbers $a,$ we have $f(a)=f(a\cdot1)=f(a)+f(1),$ from which $f(1)=0.$

For all positive rational numbers $a,$ we have $f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0,$ from which $f\left({\frac 1a}\right)=-f(a).$

For all positive integers $x$ and $y,$ suppose $\prod_{k=1}^{m}p_k^{d_k}$ and $\prod_{k=1}^{n}q_k^{e_k}$ are their respective prime factorizations. We get $\begin{align*} f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) \\ &=f(x)-f(y) && \hspace{10mm}\text{by Property 3} \\ &=f\left(\prod_{k=1}^{m}p_k^{d_k}\right)-f\left(\prod_{k=1}^{n}q_k^{e_k}\right) \\ &=\left[\sum_{k=1}^{m}f\left(p_k^{d_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{e_k}\right)\right] && \hspace{10mm}\text{by Property 1} \\ &=\left[\sum_{k=1}^{m}d_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}e_k f\left(q_k\right)\right] && \hspace{10mm}\text{by Property 1} \\ &=\left[\sum_{k=1}^{m}d_k p_k \right]-\left[\sum_{k=1}^{n}e_k q_k \right]. \end{align*}$ We apply $f$ to each fraction in the answer choices: $\begin{alignat*}{10} &\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && f\left(\frac{17^1}{2^5}\right) \quad && = \quad && [1(17)]-[5(2)] \quad && = \quad && 7 \\ &\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && f\left(\frac{11^1}{2^4}\right) \quad && = \quad && [1(11)]-[4(2)] \quad && = \quad && 3 \\ &\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && f\left(\frac{7^1}{3^2}\right) \quad && = \quad && [1(7)]-[2(3)] \quad && = \quad && 1 \\ &\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && f\left(\frac{7^1}{2^1\cdot3^1}\right) \quad && = \quad && [1(7)]-[1(2)+1(3)] \quad && = \quad && 2 \\ &\textbf{(E)} \qquad && f\left(\frac{25}{11}\right) \quad && = \quad && f\left(\frac{5^2}{11^1}\right) \quad && = \quad && [2(5)]-[1(11)] \quad && = \quad && {-}1 \end{alignat*}$ Therefore, the answer is $\boxed{\textbf{(E) }\frac{25}{11}}.$

~MRENTHUSIASM

Solution 5 (Quick, Dirty, and Frantic Last Hope)

Note that answer choices $\textbf{(A)}$ through $\textbf{(D)}$ are $\frac{\text{prime}}{\text{composite}},$ whereas $\textbf{(E)}$ is $\frac{\text{composite}}{\text{prime}}.$ Because the functional equation is related to primes, we hope that the uniqueness of answer choice $\boxed{\textbf{(E) }\frac{25}{11}}$ is enough.

~OliverA

Solution 6 (Rushed Generalization)

If f(a $\cdot$ b) = f(a) + f(b), and if f(p) = p, then f(p $\cdot$ 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 $\frac{1}{6}$ as $\frac{1}{2}$ $\cdot$ $\frac{1}{3}$ . When you get to E, you get f(25) + f( $\frac{1}{11}$ ), which is 10 - 11, which is -1. So the answer is $\boxed{\textbf{(E) }\frac{25}{11}}$

~Zeeshan12

Solution 7 (Generalized)

Note that for each of the answer choices we can multiply the fractions by their denominators to be left with only the numerator and also have the prime factorization of the denominators, then set the function equal to the numerator. This will be shown throughout the solution.

For answer choice $\text{(A) }\dfrac{17}{32}$ , we have $f\left(\dfrac{17}{32}\right)$ . Now, we can set up an equation by multiplying $\dfrac{17}{32}$ by $32$ and setting it equal to $17$ . This will give $f\left(32\cdot\dfrac{17}{32}\right)=f(17)\rightarrow f(32)+f\left(\dfrac{17}{32}\right)=f(17)$ . Our goal is to find $f\left(\dfrac{17}{32}\right)$ , therefore we must find $f(32)$ and $f(17)$ . Since $f(p)=p$ for any prime $p$ , $f(17)=17$ . Taking the prime factorization of $32$ gives $2^5$ , so $f(32)=f(2\cdot2\cdot2\cdot2\cdot2)=f(2)+f(2)+f(2)+f(2)+f(2)=5f(2)=5\cdot2=10$ . Therefore, $10+f\left(\dfrac{17}{32}\right)=17$ and $f\left(\dfrac{17}{32}\right)=7$

$\text{(B) }\dfrac{11}{16}$ : $f\left(\dfrac{11}{16}\right)\rightarrow f\left(16\cdot\dfrac{11}{16}\right)=f(11)$ $\rightarrow f(16)+f\left(\dfrac{11}{16}\right)=11\rightarrow 4f(2)+f\left(\dfrac{11}{16}\right)=11$ $\rightarrow f\left(\dfrac{11}{16}\right)=3$

$\text{(C) }\dfrac{7}{9}$ : $f\left(\dfrac{7}{9}\right)\rightarrow f\left(9\cdot\dfrac{7}{9}\right)=f(7)$ $\rightarrow f(9)+f\left(\dfrac{7}{9}\right)=7\rightarrow 2f(3)+f\left(\dfrac{7}{9}\right)=7$ $\rightarrow f\left(\dfrac{7}{9}\right)=1$

$\text{(D) }\dfrac{7}{6}$ : $f\left(\dfrac{7}{6}\right)\rightarrow f\left(6\cdot\dfrac{7}{6}\right)=f(7)$ $\rightarrow f(6)+f\left(\dfrac{7}{6}\right)=7\rightarrow f(2)+f(3)+f\left(\dfrac{7}{6}\right)=7$ $\rightarrow f\left(\dfrac{7}{6}\right)=2$

$\text{(E) }\dfrac{25}{11}$ : $f\left(\dfrac{25}{11}\right)\rightarrow f\left(11\cdot\dfrac{25}{11}\right)=f(25)$ $\rightarrow f(11)+f\left(\dfrac{25}{11}\right)=2f(5)\rightarrow 11+f\left(\dfrac{25}{11}\right)=10$ $\rightarrow f\left(\dfrac{25}{11}\right)=-1$

Therefore, the answer is $\boxed{\text{(E) }\dfrac{25}{11}}$

Note: The general strategy here was the setting up of equations to find $f(x)$ . By setting it equal to $f(a)$ where $a$ was an integer, we could take the prime factorization to find the value of $f(a)$ and also set up an equation involving $f(\text{denominator})$ because the denominator was also an integer, therefore we could take the prime factorization and find its value

~Tacos_are_yummy_1

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

https://youtu.be/8gGcj95rlWY

Video Solution by OmegaLearn (Using Functions and Manipulations)

https://youtu.be/aGv99CLzguE

~ 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

@@ Line 1: / Line 1: @@
-{{duplicate|[[2021 AMC 10A Problems#Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems#Problem 18|2021 AMC 12A #18]]}}
+{{duplicate|[[2021 AMC 10A Problems/Problem 18|2021 AMC 10A #18]] and [[2021 AMC 12A Problems/Problem 18|2021 AMC 12A #18]]}}
 ==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>. Furthermore, 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>?
+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) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad</math>
+<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 (Intuitive)==
@@ Line 57: / Line 57: @@
 ==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+....+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>A,B,C, </math> and <math>D</math>, the numerator <math>(a)</math> has less prime factors than the denominator, and so they are less likely to work. We check <math>E</math> first, and it works, therefore the answer is <math>\boxed{\textbf{(E) }\frac{25}{11}}</math>.
+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
-==Solution 4 (Generalized: Extremely Comprehensive)==
+==Solution 4 (Generalized)==
-===Results===
+We derive the following properties of <math>f:</math>
-We have the following important results:
 <ol style="margin-left: 1.5em;">
-   <li><math>f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)</math> for all positive rational numbers <math>a_k</math> and positive integers <math>n</math></li><p>
+   <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>
-  <li><math>f\left(a^n\right)=nf(a)</math> for all positive rational numbers <math>a</math> and positive integers <math>n</math></li><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><math>f(1)=0</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><math>f\left({\frac 1a}\right)=-f(a)</math> for all positive rational numbers <math>a</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>
-~MRENTHUSIASM
+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
-===Proofs===
-<ol style="margin-left: 1.5em;">
-  <li>Result 1: We can show Result 1 by induction.</li><p>
-  <li>Result 2: Since positive powers are just repeated multiplication of the base, we will use Result 1 to prove Result 2: <cmath>f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).</cmath></li><p>
-  <li>Result 3: 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> from which Result 3 is true.</li><p>
-  <li>Result 4: We have <cmath>f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.</cmath> Therefore, we get <math>f\left({\frac 1a}\right)=-f(a),</math> from which Result 4 is true.</li><p>
-</ol>
-~MRENTHUSIASM
-===Solution===
-For all positive integers <math>x</math> and <math>y,</math> suppose <math>\prod_{k=1}^{m}p_k^{e_k}</math> and <math>\prod_{k=1}^{n}q_k^{d_k}</math> are their respective prime factorizations, we have
 <cmath>\begin{align*}
-f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) && \hspace{10mm}\text{by Result 1} \\
+f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) \\
-&=f(x)-f(y) && \hspace{10mm}\text{by Result 4} \\
+&=f(x)-f(y) && \hspace{10mm}\text{by Property 3} \\
-&=f\left(\prod_{k=1}^{m}p_k^{e_k}\right)-f\left(\prod_{k=1}^{n}q_k^{d_k}\right) \\
+&=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^{e_k}\right)\right]-\left[\sum_{k=1}^{n}f\left(q_k^{d_k}\right)\right] && \hspace{10mm}\text{by Result 1} \\
+&=\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}e_k f\left(p_k\right)\right]-\left[\sum_{k=1}^{n}d_k f\left(q_k\right)\right] && \hspace{10mm}\text{by Result 2} \\
+&=\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}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right].
+&=\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 function <math>f</math> to each fraction in the answer choices:
+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 \\
@@ Line 97: / Line 84: @@
 &\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
+&\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>
@@ Line 103: / Line 90: @@
 ~MRENTHUSIASM
-==Solution 5==
+==Solution 5 (Quick, Dirty, and Frantic Last Hope)==
-The problem gives us that <math>f(p)=p.</math> If we let <math>a=p</math> and <math>b=1,</math> we get <math>f(p)=f(p)+f(1),</math> which implies <math>f(1)=0.</math> Notice that the answer choices are all fractions, which means we will have to multiply an integer by a fraction to be able to solve it. Therefore, let's try plugging in fractions and try to solve them. Note that if we plug in <math>a=p</math> and <math>b=1/p,</math> we get <math>f(1)=f(p)+f(1/p).</math> We can solve for <math>f(1/p)</math> as <math>-f(p)!</math> This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of <math>\boxed{\textbf{(E) }\frac{25}{11}}.</math>
+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
+==Solution 7 (Generalized)==
+Note that for each of the answer choices we can multiply the fractions by their denominators to be left with only the numerator and also have the prime factorization of the denominators, then set the function equal to the numerator. This will be shown throughout the solution.
+For answer choice <math>\text{(A) }\dfrac{17}{32}</math>, we have <math>f\left(\dfrac{17}{32}\right)</math>. Now, we can set up an equation by multiplying <math>\dfrac{17}{32}</math> by <math>32</math> and setting it equal to <math>17</math>. This will give <math>f\left(32\cdot\dfrac{17}{32}\right)=f(17)\rightarrow f(32)+f\left(\dfrac{17}{32}\right)=f(17)</math>. Our goal is to find <math>f\left(\dfrac{17}{32}\right)</math>, therefore we must find <math>f(32)</math> and <math>f(17)</math>. Since <math>f(p)=p</math> for any prime <math>p</math>, <math>f(17)=17</math>. Taking the prime factorization of <math>32</math> gives <math>2^5</math>, so <math>f(32)=f(2\cdot2\cdot2\cdot2\cdot2)=f(2)+f(2)+f(2)+f(2)+f(2)=5f(2)=5\cdot2=10</math>. Therefore, <math>10+f\left(\dfrac{17}{32}\right)=17</math> and <math>f\left(\dfrac{17}{32}\right)=7</math>
+<math>\text{(B) }\dfrac{11}{16}</math>: <math>f\left(\dfrac{11}{16}\right)\rightarrow f\left(16\cdot\dfrac{11}{16}\right)=f(11)</math>
+<math>\rightarrow f(16)+f\left(\dfrac{11}{16}\right)=11\rightarrow 4f(2)+f\left(\dfrac{11}{16}\right)=11</math>
+<math>\rightarrow f\left(\dfrac{11}{16}\right)=3</math>
+<math>\text{(C) }\dfrac{7}{9}</math>: <math>f\left(\dfrac{7}{9}\right)\rightarrow f\left(9\cdot\dfrac{7}{9}\right)=f(7)</math>
+<math>\rightarrow f(9)+f\left(\dfrac{7}{9}\right)=7\rightarrow 2f(3)+f\left(\dfrac{7}{9}\right)=7</math>
+<math>\rightarrow f\left(\dfrac{7}{9}\right)=1</math>
+<math>\text{(D) }\dfrac{7}{6}</math>: <math>f\left(\dfrac{7}{6}\right)\rightarrow f\left(6\cdot\dfrac{7}{6}\right)=f(7)</math>
+<math>\rightarrow f(6)+f\left(\dfrac{7}{6}\right)=7\rightarrow f(2)+f(3)+f\left(\dfrac{7}{6}\right)=7</math>
+<math>\rightarrow f\left(\dfrac{7}{6}\right)=2</math>
+<math>\text{(E) }\dfrac{25}{11}</math>: <math>f\left(\dfrac{25}{11}\right)\rightarrow f\left(11\cdot\dfrac{25}{11}\right)=f(25)</math>
+<math>\rightarrow f(11)+f\left(\dfrac{25}{11}\right)=2f(5)\rightarrow 11+f\left(\dfrac{25}{11}\right)=10</math>
+<math>\rightarrow f\left(\dfrac{25}{11}\right)=-1</math>
+Therefore, the answer is <math>\boxed{\text{(E) }\dfrac{25}{11}}</math>
+Note: The general strategy here was the setting up of equations to find <math>f(x)</math>. By setting it equal to <math>f(a)</math> where <math>a</math> was an integer, we could take the prime factorization to find the value of <math>f(a)</math> and also set up an equation involving <math>f(\text{denominator})</math> because the denominator was also an integer, therefore we could take the prime factorization and find its value
+~Tacos_are_yummy_1
 ==Video Solution by Hawk Math==
@@ Line 115: / Line 136: @@
 https://youtu.be/8gGcj95rlWY
-== Video Solution by OmegaLearn (Using Functions and manipulations) ==
+== Video Solution by OmegaLearn (Using Functions and Manipulations) ==
 https://youtu.be/aGv99CLzguE
@@ Line 124: / Line 145: @@
 ~IceMatrix
+==Video Solution (Quick and Easy)==
+https://youtu.be/NbAu_STtcvA
+~Education, the Study of Everything
 ==See also==

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 10 Problems and Solutions

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 12 Problems and Solutions

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