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! 

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.
