Find the last 3 digits of $2015^{18}+2015^{19}+2015^{20}+...+2015^{2018}$. Obvious I would find the sum of the numbers mod 100, but how would I handle this when the powers are so large? Calculating one already seemds ridiculous.

As $2015\equiv15\pmod{1000},$ we need $\displaystyle\sum_{r=18}^{2018}15^r\pmod{1000}$
As $(15^n,1000)=5^3$ for $n\ge3,$
let us find  $\displaystyle  S=\sum_{r=18-3}^{2018-3}15^r\pmod{1000/5^3}$
As $\displaystyle15\equiv-1\pmod8,S\equiv\sum_{r=18-3}^{2018-3}(-1)^r\equiv-1\pmod8$
$\displaystyle15^3S\equiv15^3\cdot-1\pmod{15^3\cdot8}\equiv-15^3\pmod{8\cdot5^3}$ as $5^3|15^3$
Now $-15^3=-225\cdot15=-3375\equiv-375\equiv1000-375\pmod{1000}$ as $S>0$
