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!

Solution 1:

\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}

Solution 2:

If you can multiply a 3-digit number by $3$ without a calculator, then you can answer the question without a calculator. Just start with $1$, multiply by $3$ $27$ times, keeping only the last three digits. $1,3,9,27,81,243,729,187$, and so on.