Difference between revisions of "2023 AMC 12A Problems/Problem 22"

Revision as of 22:01, 9 November 2023

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 $p,q$ 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 not 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

Video Solution by MOP 2024

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

~r00tsOfUnity

@@ Line 63: / Line 63: @@
 So not we get
 <cmath>\frac{1}{n}= \sum_{d\mid n}\frac{f(d)}{d}</cmath>
-Also, notice that both <math>\frac{f(d)}{d}</math> and <math>\frac{1}{n}</math> are both arithmetic functions. Applying Möbius inversion formula， we get
+Also, notice that both <math>\frac{f(d)}{d}</math> and <math>\frac{1}{n}</math> are arithmetic functions. Applying Möbius inversion formula， we get
 <cmath>\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) </cmath>
 So

2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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