Using modular arithmetic, solve the following:
Find the remainder of $(2014^{2015} \cdot 2016^{2017}) + 2018^{2019}$ when it is divided by 13.
I have no idea where to start. I've tried putting this in (mod 13) but I'm not sure where to start or even finish.
Any help would be appreciated.
You need the little Fermat's theorem (or very much patience...). This theorem guarantee that, in particular, $x^{13}\equiv x\pmod {13}$. Or more generally, $$x^{n+12k}\equiv x^n\pmod {13}$$ for any integer $k$.
Now, $2014\equiv -1\pmod{13}$, $2016\equiv 1\pmod {13}$ and $2018\equiv 3\pmod {13}$, so
$$2104^{2015}\cdot2016^{2017}+2018^{2019}\equiv -1\cdot 1+3^{168\cdot12+3}\equiv -1+3^3\equiv 0\pmod{13}$$