2023 AMC 12A Problems/Problem 22: Difference between revisions

Latest revision as of 09:20, 11 November 2025

Problem

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 $n$ . What is $f(2023)$ ?

$\textbf{(A)}~-1536\qquad\textbf{(B)}~96\qquad\textbf{(C)}~108\qquad\textbf{(D)}~116\qquad\textbf{(E)}~144$

Solution 1 (Very Thorough)

First, we note that $f(1) = 1$ , since the only divisor of $1$ is itself.

Then, let's look at $f(p)$ for $p$ a prime. We see that $\sum_{d \mid p} d \cdot f\left(\frac{p}{d}\right) = 1$ $1 \cdot f(p) + p \cdot f(1) = 1$ $f(p) = 1 - p \cdot f(1)$ $f(p) = 1-p$ Nice.

Now consider $f(p^k)$ , for $k \in \mathbb{N}$ . $\sum_{d \mid p^k} d \cdot f\left(\frac{p^k}{d}\right) = 1$ $1 \cdot f(p^k) + p \cdot f(p^{k-1}) + p^2 \cdot f(p^{k-2}) + \dotsc + p^k f(1) = 1$ .

It can be (strongly) inductively shown that $f(p^k) = f(p) = 1-p$ . Here's how.

We already showed $k=1$ works. Suppose it holds for $k = n$ , then

$1 \cdot f(p^n) + p \cdot f(p^{n-1}) + p^2 \cdot f(p^{n-2}) + \dotsc + p^n f(1) = 1 \implies f(p^m) = 1-p \; \forall \; m \leqslant n$

For $k = n+1$ , we have

$1 \cdot f(p^{n+1}) + p \cdot f(p^{n}) + p^2 \cdot f(p^{n-1}) + \dotsc + p^{n+1} f(1) = 1$ , then using $f(p^m) = 1-p \; \forall \; m \leqslant n$ , we simplify to

$1 \cdot f(p^{n+1}) + p \cdot (1-p) + p^2 \cdot (1-p) + \dotsc + p^n \cdot (1-p) + p^{n+1} f(1) = 1$ $f(p^{n+1}) + \sum_{i=1}^n p^i (1-p) + p^{n+1} = 1$ $f(p^{n+1}) + p(1 - p^n) + p^{n+1} = 1$ $f(p^{n+1}) + p = 1 \implies f(p^{n+1}) = 1-p$ .

Very nice! Now, we need to show that this function is multiplicative, i.e. $f(pq) = f(p) \cdot f(q)$ for $\textbf{distinct}$ $p,q$ prime. It's pretty standard, let's go through it quickly. $\sum_{d \mid pq} d \cdot f\left(\frac{pq}{d}\right) = 1$ $1 \cdot f(pq) + p \cdot f(q) + q \cdot f(p) + pq \cdot f(1) = 1$ Using our formulas from earlier, we have $f(pq) + p(1-q) + q(1-p) + pq = 1 \implies f(pq) = 1 - p(1-q) - q(1-p) - pq = (1-p)(1-q) = f(p) \cdot f(q)$

Great! We're almost done now. Let's actually plug in $2023 = 7 \cdot 17^2$ into the original formula. $\sum_{d \mid 2023} d \cdot f\left(\frac{2023}{d}\right) = 1$ $1 \cdot f(2023) + 7 \cdot f(17^2) + 17 \cdot f(7 \cdot 17) + 7 \cdot 17 \cdot f(17) + 17^2 \cdot f(7) + 7 \cdot 17^2 \cdot f(1) = 1$ Let's use our formulas! We know $f(7) = 1-7 = -6$ $f(17) = 1-17 = -16$ $f(7 \cdot 17) = f(7) \cdot f(17) = (-6) \cdot (-16) = 96$ $f(17^2) = f(17) = -16$

So plugging ALL that in, we have $f(2023) = 1 - \left(7 \cdot (-16) + 17 \cdot (-6) \cdot (-16) + 7 \cdot 17 \cdot (-16) + 17^2 \cdot (-6) + 7 \cdot 17^2\right)$

which, be my guest simplifying, is $\boxed{\textbf{(B)} \ 96}$

~ $\color{magenta} zoomanTV$

Solution 2

First, change the problem into an easier form. $\sum_{d\mid n}d\cdot f(\frac{n}{d} )=\sum_{d\mid n}\frac{n}{d}f(d)=1$ So now we get $\frac{1}{n}= \sum_{d\mid n}\frac{f(d)}{d}$ Also, notice that both $\frac{f(d)}{d}$ and $\frac{1}{n}$ are arithmetic functions. Applying Möbius inversion formula, we get $\frac{f(n)}{n}=\sum_{d\mid n}\frac{ \mu (d) }{\frac{n}{d} }=\frac{1}{n} \sum_{d\mid n}d\cdot \mu (d)$ So $f(n)=1-p_1-p_2-...+p_1p_2+...=(1-p_1)(1-p_2)...=\prod_{p\mid n}(1-p)$ So the answer should be $f(2023)=\prod_{p\mid 2023}(1-p)=(1-7)(1-17)=\boxed{\textbf{(B)} \ 96}$

~ZZZIIIVVV

Solution 3

From the problem, we want to find $f(2023)$ . Using the problem, we get $f(2023)+7f(289)+17f(119)+119f(17)+289f(7)+2023f(1)=1$ . By plugging in factors of $2023$ , we get $\begin{align} f(7)+7f(1)=1\\ f(17)+17f(1)=1\\ f(119)+7f(17)+17f(7)+119f(1)=1\\ f(289)+17f(17)+289f(1)=1 \end{align}$ Notice that $(4)-17(2)=f(289)$ , so $f(289)=-16$ . Similarly, notice that $(3)-17(1)=f(119)+7f(17)=-16$ . Now, substituting this all back into our equation to solve for $f(2023)$ , we get $\begin{align*} f(2023)+7f(289)+17(f(119)+7f(17))+289(f(7)+7f(1))=1\\ f(2023)+7 \cdot (-16) + 17 \cdot (-16) + 289 \cdot (1) = 1\\ f(2023)=\boxed{\textbf{(B)} \ 96} \end{align*}$ -PhunsukhWangdu

Solution 4

Consider any $n \in \Bbb N$ with prime factorization $n = \Pi_{i=1}^k p_i^{\alpha_i}$ . Thus, the equation given in this problem can be equivalently written as $\sum_{\beta_1 = 0}^{\alpha_1} \sum_{\beta_2 = 0}^{\alpha_2} \cdots \sum_{\beta_k = 0}^{\alpha_k} \Pi_{i=1}^k p_i^{\alpha_i - \beta_i} \cdot f \left( \Pi_{i=1}^k p_i^{\beta_i} \right) = 1 .$

$\noindent \textbf{Special case 1}$ : $n = 1$ .

We have $f \left( 1 \right) = 1$ .

$\noindent \textbf{Special case 2}$ : $n$ is a prime.

We have $1 \cdot f \left( n \right) + n \cdot f \left( 1 \right) = 1 .$

Thus, $f \left( n \right) = 1 - n$ .

$\noindent \textbf{Special case 3}$ : $n$ is the square of a prime, $n = p_1^2$ .

We have $1 \cdot f \left( p_1^2 \right) + p_1 \cdot f \left( p_1 \right) + p_1^2 \cdot f \left( 1 \right) = 1.$

Thus, $f \left( p_1^2 \right) = 1 - p_1$ .

$\noindent \textbf{Special case 4}$ : $n$ is the product of two distinct primes, $n = p_1 p_2$ .

We have $1 \cdot f \left( p_1 p_2 \right) + p_1 \cdot f \left( p_2 \right) + p_2 \cdot f \left( p_1 \right) + p_1 p_2 \cdot f \left( 1 \right) = 1.$

Thus, $f \left( p_1 p_2 \right) = 1 - p_1 - p_2 + p_1 p_2$ .

$\noindent \textbf{Special case 5}$ : $n$ takes the form $n = p_1^2 p_2$ , where $p_1$ and $p_2$ are two distinct primes.

We have $1 \cdot f \left( p_1^2 p_2 \right) + p_1 \cdot f \left( p_1 p_2 \right) + p_1^2 \cdot f \left( p_2 \right) + p_2 \cdot f \left( p_1^2 \right) + p_1 p_2 f \left( p_1 \right) + p_1^2 p_2 f \left( 1 \right) = 1.$

Thus, $f \left( p_1^2 p_2 \right) = 1 - p_1 - p_2 + p_1 p_2$ .

The prime factorization of 2023 is $7 \cdot 17^2$ . Therefore, $\begin{align*} f \left( 2023 \right) & = 1 - 7 - 17 + 7 \cdot 17 \\ & = \boxed{\textbf{(B) 96}}. \end{align*}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 5 (Very much overkill, but fast if you know stuff)

In the below, denote $\operatorname{id}(x)=x$ and $1(x)=1$ , and let the Dirichlet convolution of $f$ and $g$ be $f*g$ . Denote the Dirichlet inverse of $f$ as $f^{-1}$ . Also, let the Dirichlet series of $f$ evaluated at $s$ be: $DG_f(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}.$ Let $\zeta(s)$ be the Riemann zeta function.

We note that $\sum_{d|n}{d\cdot f\left(\frac{n}{d}\right)}=1$ is equivalent to $f=1*\operatorname{id}^{-1}$ . We state a well-known theorem: $DG_f(s)DG_g(s)=DG_{f*g}(s)$ . By this: $DG_{\operatorname{id}^{-1}}(s)=\frac{1}{DG_{\operatorname{id}}(s)}=\frac{1}{\zeta(s-1)}.$ We know also that: $\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$ , so: $DG_{\operatorname{id}^{-1}}(s)=\sum_{n=1}^{\infty}\frac{n\mu(n)}{n^s},$ i.e., $\operatorname{id}^{-1}(n)=n\mu(n).$ Thus: \begin{align*} f(2023)&=\left(\operatorname{id}^{-1}*1\right)(2023)\\ &=\sum_{d|2023}{n\mu(n)}\\ &=\mu(1)+7\mu(7)+17\mu(17)+7\cdot 17\mu(7\cdot 17)\\ &= 7\cdot 17-7-17+1\\ &=96 \end{align*}

Hence we choose $\boxed{(B)}$ . ~Sivin

Video Solution by MOP 2024

https://YouTube.com/watch?v=gdhVqdRhMsQ

~r00tsOfUnity

Video Solution by OmegaLearn

https://youtu.be/Trz8DEmgAtk

Video Solution

https://youtu.be/Fyd1hGGHZ8k