2021 AMC 12A Problems/Problem 18

Revision as of 00:48, 15 February 2021 by MRENTHUSIASM (talk | contribs) (Solution 4 (Comprehensive, Similar to Solution 3))
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 (but where do you get $10=11+f(\frac{25}{11})$

Looking through the solutions we can see that $f(\frac{25}{11})$ can be expressed as $f(\frac{25}{11} \cdot 11) = f(11) + f(\frac{25}{11})$ so using the prime numbers to piece together what we have we can get $10=11+f(\frac{25}{11})$, so $f(\frac{25}{11})=-1$ or $\boxed{E}$.

-Lemonie

$f(\frac{25}{11} \cdot 11) = f(25) = f(5) + f(5) = 10$

- awesomediabrine

Solution 2

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 (Deeper)

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 $\boxed{\textbf{(E)}}$.

~yofro

Solution 4 (Comprehensive, Similar to Solution 3)

We have the following important results:

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

$(2) \ f(1)=0$

$(3) \ f\left({\frac 1a}\right)=-f(a) \text{ for all positive rational numbers } a$

Proofs

Result $(1)$ can be shown by induction.

Result $(2):$ For all positive rational numbers $a,$ we have \[f(a)=f(a\cdot1)=f(a)+f(1).\] Therefore, we get $f(1)=0.$

Result $(3):$ For all positive rational numbers $a,$

~MRENTHUSIASM

Video Solution by Hawk Math

https://www.youtube.com/watch?v=dvlTA8Ncp58

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

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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