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2021 AMC 12A Problems/Problem 18

Revision as of 07:30, 16 June 2021 by MRENTHUSIASM (talk | contribs) (→‎Solution 4 (Generalized: Extremely Comprehensive))
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 $a=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, $f(a)=a_1p_1+a_2p_2+....+a_kp_k$. We simply need this to be greater than what we have for $f(b)$. Notice that for answer choices $A,B,C,$ and $D$, the numerator $(a)$ has less prime factors than the denominator, and so they are less likely to work. We check $E$ first, and it works, therefore the answer is $\boxed{\textbf{(E) }\frac{25}{11}}$.

~yofro

Solution 4 (Generalized: Extremely Comprehensive)

Results

We have the following important results:

  1. $f\left(\prod_{k=1}^{n}a_k\right)=\sum_{k=1}^{n}f(a_k)$ for all positive rational numbers $a_k$ and positive integers $n$

  2. $f\left(a^n\right)=nf(a)$ for all positive rational numbers $a$ and positive integers $n$

  3. $f(1)=0$

  4. $f\left({\frac 1a}\right)=-f(a)$ for all positive rational numbers $a$

~MRENTHUSIASM

Proofs

  1. Result 1: We can show Result 1 by induction.

  2. Result 2: Since positive powers are just repeated multiplication of the base, we will use Result 1 to prove Result 2: \[f\left(a^n\right)=f\left(\prod_{k=1}^{n}a\right)=\sum_{k=1}^{n}f(a)=nf(a).\]

  3. Result 3: For all positive rational numbers $a,$ we have \[f(a)=f(a\cdot1)=f(a)+f(1).\] Therefore, we get $f(1)=0,$ from which Result 3 is true.

  4. Result 4: We have \[f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.\] Therefore, we get $f\left({\frac 1a}\right)=-f(a),$ from which Result 4 is true.

~MRENTHUSIASM

Solution

For all positive integers $x$ and $y,$ suppose $\prod_{k=1}^{m}p_k^{e_k}$ and $\prod_{k=1}^{n}q_k^{d_k}$ are their respective prime factorizations, we have \begin{align*} f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) && \hspace{10mm}\text{by Result 1} \\ &=f(x)-f(y) && \hspace{10mm}\text{by Result 4} \\ &=f\left(\prod_{k=1}^{m}p_k^{e_k}\right)-f\left(\prod_{k=1}^{n}q_k^{d_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}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}e_k p_k \right]-\left[\sum_{k=1}^{n}d_k q_k \right]. \end{align*} We apply function $f$ 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 (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

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