Ta có: $\varphi(100)=40$, nên $a^{40}\equiv 1$ (mod $100$), $\forall (a,100)=1$.
$\Rightarrow a^{40k+r}\equiv a^r$ (mod $100$), $\forall (a,100)=1$.
Mà $10^{2010}\equiv0$ (mod $40$) $\Rightarrow 23^{10^{2010}}\equiv23^0\equiv 1$ (mod $100$).
$9^{4}=6561\equiv 1$ (mod $40$) $\Rightarrow 9^{1996}\equiv1^{499}\equiv1$ (mod $40$).
$\Rightarrow 13^{9^{1996}}\equiv13^1\equiv 13$ (mod $100$).
$\Rightarrow 23^{10^{2010}}+13^{9^{1996}}\equiv 14$ (mod $100$)