2021 AMC 12A Spring Problem 18

Attempt Problem 18 of the 2021 AMC 12A Spring below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

Let ff be a function defined on the set of positive rational numbers with the property that f(a⋅b)=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}
Answer: E
Concepts:functional equationprime factorization
Small hint:
The rule makes ff additive over prime factors, and f(1p)=−f(p)=−pf(\frac{1}{p}) = -f(p) = -p
Big hint:
For x=∏pep,x = \prod p^{e_p}, compute f(x)=∑ep pf(x) = \sum e_p\, p (with negative exponents for denominators)
Solution:
The functional equation makes ff completely additive: for x=∏pep,x = \prod p^{e_p}, we have f(x)=∑ep f(p)=∑ep p,f(x) = \sum e_p\, f(p) = \sum e_p\, p, where a prime in the denominator contributes a negative exponent (since f(1p)=−pf(\frac{1}{p}) = -p).
Evaluating: f ⁣(1732)=17−5⋅2=7,f\!\left(\tfrac{17}{32}\right) = 17 - 5\cdot 2 = 7, f ⁣(1116)=11−4⋅2=3,f\!\left(\tfrac{11}{16}\right) = 11 - 4\cdot 2 = 3, f ⁣(79)=7−2⋅3=1,f\!\left(\tfrac{7}{9}\right) = 7 - 2\cdot 3 = 1, f ⁣(76)=7−2−3=2,f\!\left(\tfrac{7}{6}\right) = 7 - 2 - 3 = 2, and f ⁣(2511)=2⋅5−11=−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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