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

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

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 (Most Comprehensive, Similar to Solution 3)

We have the following important results:

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

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

$(3) \ f(1)=0$

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

Proofs

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

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

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

Result $(4):$ For all positive rational numbers $a,$ we have $$f(a)+f\left(\frac1a\right)=f\left(a\cdot\frac1a\right)=f(1)=0.$$ It follows that $f\left({\frac 1a}\right)=-f(a),$ and result $(4)$ is true.

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 prime factorizations, respectively, we have \begin{align*} f\left(\frac xy\right)&=f(x)+f\left(\frac 1y\right) \\ &=f(x)-f(y) \\ &=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] \\ &=\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] \\ &=\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 $f$ on each fraction in the choices:

$$\begin{array}{cccccccc} \textbf{(A) } & f\left(\frac{17}{32}\right) & = & f\left(\frac{17^1}{2^5}\right) & = & [1(17)]-[5(2)] & = & 7 \\ [2ex] \textbf{(B) } & f\left(\frac{11}{16}\right) & = & f\left(\frac{11^1}{2^4}\right) & = & [1(11)]-[4(2)] & = & 3 \\ [2ex] \textbf{(C) } & f\left(\frac{7}{9}\right) & = & f\left(\frac{7^1}{3^2}\right) & = & [1(7)]-[2(3)] & = & 1 \\ [2ex] \textbf{(D) } & f\left(\frac{7}{6}\right) & = & f\left(\frac{7^1}{2^1\cdot3^1}\right) & = & [1(7)]-[1(2)+1(3)] & = & 2 \\ [2ex] \textbf{(E) } & f\left(\frac{25}{11}\right) & = & f\left(\frac{5^2}{11^1}\right) & = & [2(5)]-[1(11)] & = & -1. \end{array}$$ 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 E.

~ pi_is_3.14