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2023 AMC 12A Problem 22

Problem 22 of 25HarderAlgebraNumber Theory

Let ff be the unique function defined on the positive integers such that dndf(nd)=1 \sum_{d\mid n} d\cdot f\left(\frac{n}{d}\right)=1 for all positive integers n,n, where the sum is taken over all positive divisors of n.n. What is f(2023)?f(2023)?

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Solution

Setting n=1n=1 gives f(1)=1.f(1)=1. For a prime p,p, n=pn=p gives f(p)+pf(1)=1,f(p)+p\cdot f(1)=1, so f(p)=1p.f(p)=1-p. For n=p2,n=p^2, f(p2)+pf(p)+p2f(1)=1f(p^2)+p\,f(p)+p^2 f(1)=1 gives f(p2)=1p.f(p^2)=1-p. Since the defining relation is a Dirichlet convolution of multiplicative functions, ff is multiplicative. With 2023=7172,2023=7\cdot 17^2, f(2023)=f(7)f(172)=(17)(117)=(6)(16)=96. \begin{gathered} f(2023)=f(7)\cdot f(17^2)\\ {}=(1-7)(1-17)\\ {}=(-6)(-16)\\ {}=96. \end{gathered} Thus, the correct answer is B.

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Concepts: prime factorization · recursion

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.