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2021 AMC 10A Problem 18

Problem 18 of 25IntermediateAlgebraNumber Theory

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?

Answer choices

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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)=1752=7,f(1116)=1142=3,f(79)=723=1,f(76)=723=2,f(2511)=2511=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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Concepts: functional equation · prime factorization

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