<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] We have that $$\frac {3^{11}-1}{2} = \frac {3-1}{2}(3^{10} + 3^9 + \cdots + 3^2+3+1) \tag {1} $$
We know that $3^2 =9 \equiv 0 \mod 9$ and similarly $3^3 =3 (3^2) \equiv 0 \mod 9$ and so on. Thus, $(1)$ reduces to, $$1 (0 + 0 + 0 \cdots 0 + 3 +1 ) \equiv 4 \mod 9$$
Hope it helps. 
</s>