Converting repeating decimal in base b to a fraction the same base The repeating decimal .36666... in base 8 can be written in a fraction in base 8.
I understand simple patterns such as 1/9 in base 10 is .1111.... so 1/7 in base 8 is .1111.
But I'm not too sure how to convert this decimal in this base to the fraction in the same base.  
 A: \begin{align}
0.3\bar{6}_8
&= \frac{3}{8} + 6\left(\frac{1}{8^2}+\frac{1}{8^3} + \cdots\right)\\
&= \frac{3}{8} + \frac{6}{8^2}\left(1+\frac{1}{8}+  \frac{1}{8^2} +\cdots\right)\\
&= \frac{3}{8} + \frac{6}{8^2}\frac{1}{1-(1/8)} & \text{geometric series}\\
&= \frac{3}{8} + \frac{3}{28}\\
&= \frac{27}{56}\\
&= \frac{33_8}{70_8}.
\end{align}
A: You could just do all your thinking in base 8. To save writing all the subscripts in the following computations I'll omit the base 8 designation. Legal digits are $0$ through $7$. It's a little mindbending, but only because we're used to base 10.
Let $x=0.3666\ldots$. Then 
$$
10x = 3.666\ldots = 3 + 6/7 = (25 + 6)/7 = 33/7
$$
so $x=33/70$. 
I used the facts that multiplying by 10 just shifts the "decimal" point, $3 \times 7 = 25$ and $6/7 = 0.66\ldots$.
A: We use subscript $8$ to denote base $8$ numbers, and all other numbers are base $10$. Take $x = 0.3\overline{6}_8$. Then $7x = 8x - x = 3.\overline{6}_8 - 0.3\overline{6}_8 = 3.3_8 = 27/8$. It follows that $x = 27/56 = 33_8/70_8$.
A: We compute directly in base $8$ as in base $10$:
Set $y=0.3666\dots$, $x=0.666\dots$. Then 
\begin{align*}8y&=3+0.666\dots=3+x\\
8x&=6.666\dots=6+x
\end{align*} 
Thus 
$$7x=6,\quad\text{so}\enspace x=\frac 67\enspace\text{and}\enspace 8y=3+\frac 67=\frac{3\times 7+6}7=\frac{25+6}7 =\frac{33}7,$$
because in base $8$, $\;3\times 7=25,\;5+6=13$, so that  finally $\; y=\dfrac{33}{8\times7}=\color{red}{\dfrac{33}{70}}$.
