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

(Solution 1: Merged the thoughts of the contributors and made the solution clearer. Let me know if you have other suggestions. I maintained the thoughts of the original solution.)
m (Solution 3 (Generalized))
 
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&=f\left(\frac{25}{11}\right)+11.
 
&=f\left(\frac{25}{11}\right)+11.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
Also, we have
+
On the other hand, we have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
f(25)&=f(5\cdot5) \\
 
f(25)&=f(5\cdot5) \\
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<u><b>Remark</b></u>
 
<u><b>Remark</b></u>
  
Similarly, we can calculate the outputs of <math>f</math> at the inputs of the other choices:
+
Similarly, we can find the outputs of <math>f</math> at the inputs of the other answer choices:
 
<cmath>\begin{alignat*}{10}
 
<cmath>\begin{alignat*}{10}
&\textbf{(A)} && \qquad f\left(\frac{17}{32}\right) \quad && =\quad && 7 \\
+
&\textbf{(A)} \qquad && f\left(\frac{17}{32}\right) \quad && = \quad && 7 \\  
&\textbf{(B)} && \qquad f\left(\frac{11}{16}\right) \quad && =\quad && 3 \\
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&\textbf{(B)} \qquad && f\left(\frac{11}{16}\right) \quad && = \quad && 3 \\  
&\textbf{(C)} && \qquad f\left(\frac{7}{9}\right) \quad && =\quad && 1 \\
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&\textbf{(C)} \qquad && f\left(\frac{7}{9}\right) \quad && = \quad && 1 \\  
&\textbf{(D)} && \qquad f\left(\frac{7}{6}\right) \quad && =\quad && 2
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&\textbf{(D)} \qquad && f\left(\frac{7}{6}\right) \quad && = \quad && 2
 
\end{alignat*}</cmath>
 
\end{alignat*}</cmath>
Alternatively, refer to Solutions 2 and 4 for the full calculations.
+
Alternatively, refer to Solutions 2 and 4 for the full processes.
  
 
~Lemonie ~awesomediabrine ~MRENTHUSIASM
 
~Lemonie ~awesomediabrine ~MRENTHUSIASM
  
==Solution 2==
+
==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>
 
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>
 
Subtracting <math>f(p)</math> from both sides gives <math>0 = f(1).</math>
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~JHawk0224 ~awesomediabrine
 
~JHawk0224 ~awesomediabrine
  
==Solution 3 (Deeper)==
+
==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
 
~yofro
  
==Solution 4 (Extremely Comprehensive, Similar to Solution 3)==
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==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><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>
 
  <li><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>
 
</ol>
 
 
 
~MRENTHUSIASM
 
 
 
===Proofs===
 
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
   <li>Result 1: We can show Result 1 by induction.</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>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>
+
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>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>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>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>
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   <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>
 
</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
~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*}
 
<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>
 
\end{align*}</cmath>
 
+
We apply <math>f</math> to each fraction in the answer choices:
We apply function <math>f</math> to each fraction in the choices:
 
 
 
 
<cmath>\begin{alignat*}{10}
 
<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{(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 \\  
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==Solution 5==
 
==Solution 5==
The problem gives us that f(p)=p. If we let a=p and b=1, we get f(p)=f(p)+f(1), which implies f(1)=0. 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 a=p and b=1/p, we get f(1)=f(p)+f(1/p). We can solve for f(1/p) as -f(p)! 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>..
+
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>
  
 
==Video Solution by Hawk Math==
 
==Video Solution by Hawk Math==
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https://youtu.be/8gGcj95rlWY
 
https://youtu.be/8gGcj95rlWY
  
== Video Solution by OmegaLearn (Using Functions and manipulations) ==
+
== Video Solution by OmegaLearn (Using Functions and Manipulations) ==
 
https://youtu.be/aGv99CLzguE
 
https://youtu.be/aGv99CLzguE
  

Latest revision as of 13:35, 22 September 2021

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

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 $b$. Furthermore, suppose that $f$ also has the property that $f(p)=p$ for every prime number $p$. For which of the following numbers $x$ is $f(x) < 0$?

$\textbf{(A) }\frac{17}{32}\qquad\textbf{(B) }\frac{11}{16}\qquad\textbf{(C) }\frac{7}{9}\qquad\textbf{(D) }\frac{7}{6}\qquad\textbf{(E) }\frac{25}{11}\qquad$

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 $f:$

  1. 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 $n.$

  2. For all positive rational numbers $a,$ we have \[f(a)=f(a\cdot1)=f(a)+f(1),\] from which $f(1)=0.$
  3. 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

The problem gives us that $f(p)=p.$ If we let $a=p$ and $b=1,$ we get $f(p)=f(p)+f(1),$ which implies $f(1)=0.$ 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 $a=p$ and $b=1/p,$ we get $f(1)=f(p)+f(1/p).$ We can solve for $f(1/p)$ as $-f(p)!$ This gives us the information we need to solve the problem. Testing out the answer choices gives us the answer of $\boxed{\textbf{(E) }\frac{25}{11}}.$

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

See also

2021 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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