<s>[INST] 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. 
 [/INST] $3^{11}$ is divisible by $9$.
So, $3^{11}  - 1$, when divided by $9$, will give a remainder of $8$.
Division equation- 
$a=bq+r$
$a= 3^{11}  - 1$
$b= 9$
$r= 8$
$3^{11}  - 1 = 9q + 8$
The $q$ given here will be even since $3^{11} - 1$ is even, $r$ is even and so $9q$ must also be even. Hence, $q$ is of the form $2k$ for some natural number $k$.
$3^{11}  - 1 = 18k + 8$
Divide both sides by $2.$ 
$3^{11}  - 1 = 18k + 8$
$(3^{11} - 1)/2 = 9k + 4$
Hence, the remainder is $4$ when $(3^{11} - 1)/2$ is divided by $9$.
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