<s>[INST] Pattern to last three digits of power of $3$? I'm wondering if there is a pattern to the last three digits of a a power of $3$? I need to find out the last three digits of $3^{27}$, without a calculator.
I've tried to find a pattern but can not see one? Am I missing something?
Thanks for your help in advance! 
 [/INST] \begin{align}
3^{27}=3(3^{26})=3(9^{13})& =3(10-1)^{13} \\
& \equiv 3((-1)^{13}+13(-1)^{12}(10)+\binom{13}{2}(-1)^{11}(10^2)) \pmod{1000} \\
& \equiv 3(-1+130-7800) \pmod{1000} \\
& \equiv 987 \pmod{1000} \\
\end{align}
Edit: The same method (using binomial theorem) can easily be applied to $3^n$, even for large $n$.
\begin{align}
3^{2n}=9^n & =(10-1)^n \\
& \equiv (-1)^n+n(-1)^{n-1}(10)+\binom{n}{2}(-1)^{n-2}(10^2)) \pmod{1000} \\
& \equiv (-1)^n(1-10n+100\binom{n}{2}) \pmod{1000} \\
\end{align}
\begin{align}
3^{2n+1}=3(3^{2n}) \equiv 3(-1)^n(1-10n+100\binom{n}{2}) \pmod{1000} \\
\end{align}
</s>