jmc

Since the tens digits don't matter to us, the problem is the same as finding the units digit of $3^{19}\cdot 9^{13}=3^{19}\cdot 3^{26}=3^{45}$. Now we go hunting for the pattern of units digits of powers of 3: $$\begin{array}{rl}{3}^1& =3\\ 3^2& =9\\ 3^3& =27\\ 3^4& =81\\ 3^5& =243\end{array}$$ The cycle of units digits is 4 long. Since $45÷4\equiv 1(\mathrm{mod}4)$, we know that $3^{45}$ has the same units digit as $3^1=\boxed{3}$.