Question

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

$-1536$

$96$

$108$

$116$

$144$

Answer: B

Concepts:prime factorization recursion

Difficulty rating: 2270

Solution:

Setting $n=1$ gives $f(1)=1.$ For a prime $p,$ $n=p$ gives $f(p)+p\cdot f(1)=1,$ so $f(p)=1-p.$ For $n=p^2,$ $f(p^2)+p\,f(p)+p^2 f(1)=1$ gives $f(p^2)=1-p.$

Since the defining relation is a Dirichlet convolution of multiplicative functions, $f$ is multiplicative. With $2023=7\cdot 17^2,$ $\begin{gathered} f(2023)=f(7)\cdot f(17^2)\\ {}=(1-7)(1-17)\\ {}=(-6)(-16)\\ {}=96. \end{gathered}$

Thus, the correct answer is B.

2023 AMC 12A Problem 22

Answer: B

Difficulty rating: 2270

Solution:

Problem 22 in Other Years