2021 AMC 10A Spring Problem 18

Below is the video solution and professionally curated solution for Problem 18 of the 2021 AMC 10A Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 10A Spring solutions, or check the answer key.

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Concepts:functional equationprime factorization

Difficulty rating: 1280

18.

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

1732\dfrac{17}{32}

1116\dfrac{11}{16}

79\dfrac{7}{9}

76\dfrac{7}{6}

2511\dfrac{25}{11}

Video solution:
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Written solution:

Note that for any number of the form pep^e where pp is prime, f(pe)=ef(p)=ep. f(p^e) = ef(p) = ep. This can be seen by applying the function property multiple times.

Also note that f(a)=f(abb)=f(ab)+f(b), \begin{align*}f(a) &= f\left(\dfrac{a}{b} \cdot b\right) \\&= f\left(\dfrac{a}{b}\right) + f(b),\end{align*} which gives us f(ab)=f(a)f(b). f\left(\dfrac{a}{b}\right) = f(a) - f(b).

We can now calculate each answer choice one-by-one.

f(1732)=f(17)f(32) f\left(\dfrac{17}{32}\right) = f(17) - f(32) =1752=7. = 17 - 5 \cdot 2 = 7.

f(1116)=f(11)f(16)=1142=3 \begin{gather*} f\left(\dfrac{11}{16}\right) = f(11) - f(16) \\ = 11 - 4 \cdot 2 = 3 \end{gather*}

f(79)=f(7)f(9)=723=1 \begin{gather*} f\left(\dfrac{7}{9}\right) = f(7) - f(9) \\ = 7 - 2 \cdot 3 = 1 \end{gather*}

f(76)=f(7)f(6)=7f(2)f(3)=2 \begin{gather*} f\left(\dfrac{7}{6}\right) = f(7) - f(6) \\ = 7 - f(2) - f(3) = 2 \end{gather*}

f(2511)=f(25)f(11)=2511=1 \begin{gather*} f\left(\dfrac{25}{11}\right) = f(25) - f(11) \\ = 2 \cdot 5 - 11 = -1 \end{gather*}

Thus, E is the correct answer.

Problem 18 in Other Years