2021 AMC 10A Spring Problem 18

Attempt Problem 18 of the 2021 AMC 10A Spring below, then check your answer against the video solution and professionally curated solution 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.

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) < 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:
First determine f(pe)f(p^e) and f(ab)f(\frac{a}{b})
Big hint:
Evaluate the choices by prime factorization
Video solution:
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Written solution:
Repeated use of the functional equation gives f(pe)=ef(p)=epf(p^e)=ef(p)=ep for every prime pp and positive integer e.e. Also, f(a)=f(ab)+f(b),f(a)=f\left(\frac ab\right)+f(b), so f(ab)=f(a)−f(b).f(\frac{a}{b})=f(a)-f(b).
Evaluating the choices by prime factorization, f(1732)=17−5⋅2=7,f(1116)=11−4⋅2=3,f(79)=7−2⋅3=1,f(76)=7−2−3=2,f(2511)=2⋅5−11=−1. \begin{aligned} f(\frac{17}{32})&=17-5\cdot2=7,\\ f(\frac{11}{16})&=11-4\cdot2=3,\\ f(\frac{7}{9})&=7-2\cdot3=1,\\ f(\frac{7}{6})&=7-2-3=2,\\ f(\frac{25}{11})&=2\cdot5-11=-1. \end{aligned} Only the final value is negative.
Thus, E is the correct answer.
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