smc

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 $k\in \mathbb{N}$. $$\sum _{d\mid p^k}d\cdot f\left(\frac{p^k}{d}\right)=1$$ $$1\cdot f(p^k)+p\cdot f(p^{k-1})+p^2\cdot f(p^{k-2})+\dots +p^{k}f(1)=1$$. 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 $$1\cdot f(p^n)+p\cdot f(p^{n-1})+p^2\cdot f(p^{n-2})+\dots +p^{n}f(1)=1⟹f(p^m)=1-p\forall m⩽n$$ For $k=n+1$, we have $$1\cdot f(p^{n+1})+p\cdot f(p^n)+p^2\cdot f(p^{n-1})+\dots +p^{n+1}f(1)=1$$, then using $f(p^m)=1-p\forall m⩽n$, we simplify to $$1\cdot f(p^{n+1})+p\cdot (1-p)+p^2\cdot (1-p)+\dots +p^n\cdot (1-p)+p^{n+1}f(1)=1$$ $$f(p^{n+1})+\sum _{i=1}^n{p}^i(1-p)+p^{n+1}=1$$ $$f(p^{n+1})+p(1-p^n)+p^{n+1}=1$$ $$f(p^{n+1})+p=1⟹f(p^{n+1})=1-p$$. Very nice! Now, we need to show that this function is multiplicative, i.e. $f(pq)=f(p)\cdot f(q)$ for $\text{distinct}$ $p,q$ 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 $$f(pq)+p(1-q)+q(1-p)+pq=1⟹f(pq)=1-p(1-q)-q(1-p)-pq=(1-p)(1-q)=f(p)\cdot f(q)$$ 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 $\boxed{96}$