Find the units digit of $(123^{33} + 27^{54}) \times (148^{326} + 2019^{2016})$.
It's equal to $(3^{33}+7^{54})(8^{326}+(-1)^{2016})\pmod{10}$.
$3,7,8$ all cycle modulo $10$ and $(-1)^{2016}$ is trivial to deal with.