jmc

We start by generating powers of 9 modulo 17. Note that we can generate $9^{2k}$ from $9^k$ by squaring $9^k$. We get $$\begin{array}{rl}{9}^1& \equiv 9(\mathrm{mod}17)\\ 9^2& \equiv 13(\mathrm{mod}17)\\ 9^4& \equiv 16(\mathrm{mod}17)\\ 9^8& \equiv 1(\mathrm{mod}17).\end{array}$$Since $9^8\equiv 1$ modulo 17, we have $$\begin{array}{rl}{9}^{2010}& \equiv 9^2{9}^{2008}\\ & \equiv 9^2(9^8{)}^{251}\\ & \equiv 9^2{1}^{251}\\ & \equiv 9^2\\ & \equiv \boxed{13}(\mathrm{mod}17).\end{array}$$