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.
 A: Note that the last 3 digit for $2015^2$ are the same for $15^2=225$ and then they repeat periodically


*

*$15^3=...375$

*$15^4=...625$

*$15^5=...375$

*...


then
$$2015^{18}+2015^{19}+2015^{20}+…+2015^{2018}\equiv1001\cdot625+1000\cdot 375 \pmod{1000}\equiv \Large{625}$$
A: Consider that since $2015$ to any power will have a final digit of $5$, since the last digit of $2015$ is $5$. (You can check the first few of these using a calculator)
Also consider that you have an even number of elements. Anything ending in $5$ multiplied by an even number will have a last digit of $0$. ($5 \times 2 = 10$, $15 \times 3 = 45$, $25 \times 4 = 100$). You can therefore conclude that the last digit will be a $0$.
You should be able to make conclusions about the next two digits using a similar method of generalizations.
A: Modulo $1000$,  $15^3=15^5$; so, for $k\geq 3$, $$15^k=\begin{cases}15^3,&\ k\ \text{ odd}\\ \ \\ 15^4,&\ k\ \text{ even}\end{cases}$$
Then, still modulo $1000$, 
$$
S=\sum_{k=0}^{2000} 2015^{18+k}=\sum_{k=0}^{2000} 2015^{k+2}
=15^{2}+1000\times15^3+1000\times 15^4=15^2=625. 
$$
So the last three digits are 625.
A: We need to work modulo $1000 = 2^3 5^3$.  Since $2015$ is divisible by $5$, the sum  is congruent to $0$ modulo $5^3.$  Reducing the sum modulo $2^3$ yields
$$1 -1 + 1 - 1 +\cdots -1+1 \equiv 1 \pmod{8},$$
since $2015 \equiv 15 \equiv -1 \pmod{8}$. So you need to use Chinese remainder theorem to solve the system
$$x\equiv 0 \pmod{125}$$
$$x\equiv 1 \pmod{8}$$
which yields $x \equiv 625 \pmod{1000}.$
A: 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 $S>0$
