Let $n$ be a positive integer greater than or equal to $3$. Let $a,b$ be integers such that $ab$ is invertible modulo $n$ and $(ab)^{-1}\equiv 2\pmod n$. Given $a+b$ is invertible, what is the remainder when $(a+b)^{-1}(a^{-1}+b^{-1})$ is divided by $n$?
I notice that the expression $(a+b)^{-1}(a^{-1}+b^{-1})$ can be rewritten as $\frac{1}{a+b}\cdot \frac{a+b}{ab}$ by using the fact that $x^{-1}$ means the same as $\frac{1}{x}$. This simplifies to $\frac{1}{ab}$, which is exactly what was given as $(ab)^{-1}$. So the answer is $2$, since $(ab)^{-1}\equiv 2\pmod n$. 