Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:

Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$

I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know where to go next. Anyway, this is one of my many attempts to solve this question, and most of them ends with a complicated solution. I don't want to use modular arithmetic for this question. A hint or anything will help me. 

The tool you need is modular arithmetic.
$3^{11}$ is congruent to $0$ mod $9$
$3^{11} - 1$ is congruent to $8$ mod $9$
The only hard part is the division. But, fortunately, you can uniquely divide by $2$ mod $9$:
$0*2 = 0,
1*2 = 2,
2*2 = 4,
3*2 = 6,
4*2 = 8,
5*2 = 1,
6*2 = 3,
7*2 = 5,$ and
$8*2 = 7$
so $8/2 = 4$ modulo $9$, thus the remainder is $4$.
