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! 
 A: There will be a pattern to the last three digits of a power of 3, in general. However, that pattern may not necessarily show itself within the first 27 terms.
However, here's something you can do instead to solve your problem:
$$\begin{align}
\text{ last 3 digits of } 3^{27} &= \text{ last 3 digits of } (3^3)^9\\
&= \text{ last 3 digits of } 27^9\\
&= \text{ last 3 digits of } (27^3)^3\\
&= \text{ last 3 digits of } 19683^3\\
&= \text{ last 3 digits of } 683^3\\
&= \text{ last 3 digits of } 318611987\\
&= 987
\end{align}
$$
Not the most elegant solution, but it does reduce the number (and difficulty) of the multiplications required to solve the problem.
A: \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}
A: Exponentiation by squaring reduces the number of multiplications from 26 to 7:
First square repeatedly to create powers by powers of 2:
$$ 3^1=3 \qquad 3^2=9 \qquad 3^4=81 \qquad
3^8 \equiv 561 \pmod{1000} \qquad
3^{16} \equiv 721 \pmod{1000} $$
Then, since $27=1+2+8+16$,
$$ 3^3=3^1 3^2 = 27 \qquad
3^{11} = 3^3 3^8 \equiv 147 \pmod{1000} \qquad
3^{27} = 3^{11} 3^{16} \equiv 987 \pmod{1000} $$

We can even be smarter, as gt6989b noted in a comment, and use $3^{27} = ((3^3)^3)^3$ and get down to 6 multiplications:
$$ 3^2 = 9 \qquad 3^3 = 27 $$
$$ (3^3)^2 = 729 \qquad 3^9 = (3^3)^3 \equiv 683 $$
$$ (3^9)^2 \equiv 489 \qquad 3^{27} = (3^9)^3 \equiv 987 $$
But in general the trouble of looking for such tricks is not really worth it over just using plain squaring.
A: 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. 
A: Since $1000 = 2^3 5^3$, and those factors are relatively prime, you can determine $3^n \mod 1000$ by determining $3^n \pmod 8$ and $3^n \pmod {125}$,
$3^n \pmod 8$ is easy, as $3^2 = 9 \equiv 1 \pmod 8$.
$3^n \pmod {125}$ is tedious. Work with smaller powers of $5$ as exponent first:
$3^4 \equiv 1 \pmod 5$. The exponent must divide $\phi(5) = 4$.
$3^{20} \equiv 1 \pmod{25}$. The exponent must divide $\phi(25) = 20$ and be a multiple of $4$ from the above.
$3^{20} \equiv 26 \pmod {125}$. I cheated and used a calculator.  But $3^{100} \equiv 1 \pmod {125}$. The exponent must divide $\phi(125) = 100$ and be a multiple of $20$.
So, $3^{27} \equiv 3 \pmod 8$ and $3^{27} \equiv 26\cdot3^7 \pmod {125}$.
$3^5 = 243 \equiv -7 \pmod {125}$. $3^7 \equiv -7 \cdot 9 \equiv -63 \equiv 62 \pmod {125}$. So $3^{27} \equiv 26 \cdot 62 \pmod {125}$.
Use the Chinese Remainder Theorem.
