jmc

We may begin by noting that $17\cdot 5=85\equiv 2(\mathrm{mod}83)$. However, we are looking for $n$ such that $17\cdot n\equiv 1(\mathrm{mod}83)$.

Note that $2\cdot 42=84\equiv 1(\mathrm{mod}83)$. Therefore, $$\begin{array}{rl}17\cdot 5\cdot 42& \equiv 2\cdot 42\\ & \equiv 1(\mathrm{mod}83),\end{array}$$which tells us that $17$ and $5\cdot 42$ are each other's inverses modulo $83$. We can evaluate $5\cdot 42=210$, but this is not in the range $0$ to $82$, so we take its residue $(\mathrm{mod}83)$, which is $44$.

Therefore, $17^{-1}\equiv \boxed{44}(\mathrm{mod}83)$.

We may check our answer: $17\cdot 44=748=9\cdot 83+1\equiv 1(\mathrm{mod}83)$, so our answer is correct.