2021 AMC 12A Spring Problem 18

Below is the professionally curated solution for Problem 18 of the 2021 AMC 12A Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Spring solutions, or check the answer key.

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Concepts:functional equationprime factorization

Difficulty rating: 1950

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) \lt 0?

1732\dfrac{17}{32}

1116\dfrac{11}{16}

79\dfrac{7}{9}

76\dfrac{7}{6}

2511\dfrac{25}{11}

Solution:

The functional equation makes ff completely additive: for x=pep,x = \prod p^{e_p}, we have f(x)=epf(p)=epp,f(x) = \sum e_p\, f(p) = \sum e_p\, p, where a prime in the denominator contributes a negative exponent (since f(1/p)=pf(1/p) = -p).

Evaluating: f ⁣(1732)=1752=7,f\!\left(\tfrac{17}{32}\right) = 17 - 5\cdot 2 = 7, f ⁣(1116)=1142=3,f\!\left(\tfrac{11}{16}\right) = 11 - 4\cdot 2 = 3, f ⁣(79)=723=1,f\!\left(\tfrac{7}{9}\right) = 7 - 2\cdot 3 = 1, f ⁣(76)=723=2,f\!\left(\tfrac{7}{6}\right) = 7 - 2 - 3 = 2, and f ⁣(2511)=2511=1.f\!\left(\tfrac{25}{11}\right) = 2\cdot 5 - 11 = -1. Only the last is negative.

Thus, the correct answer is E.

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