Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is

Write each of the following as a
  decimal and a fraction:
(iii) $66\frac{2}{3}$%
(iv) $16\frac{2}{3}$%

For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{6} = 0.\bar{6}$ and a fraction $66\frac{2}{3}\% = \frac{66*3 + 2}{3*100}\% = \frac{200}{3}\times\frac{1}{100} =\frac{2}{3}$.
For (iv), I follow a similar path to establish that the fraction is $\frac{1}{6}$ and the decimal is $0.1\bar{6}$.
What I don't understand is a part of the model answer for this exercise. They say

$66\frac{2}{3}\% = 66.\bar{6}\% = 0.\bar{6} = \frac{6}{9} = \frac{2}{3}$

and

$16\frac{2}{3}\% = 16.\bar{6}\% = 0.1\bar{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}$

I did not do much work with recurring decimals before and I don't know how to justify the 9 and 90 in the denominator. 
Could you please explain?
 A: Slightly informally, you just multiply to make the repeating decimals subtract out.
$\begin{align} 10x &=6.\bar{6} \\x &=0.\bar{6} \\9x &=6 \\ x&=\frac{6}{9} \end{align}$ 
The case for $0.1\bar{6}$ is similar:
$\begin{align} 100x &=16.\bar{6} \\  10x &=1.\bar{6} \\  90x &=15 \\  x&=\frac{15}{90} \end{align}$
A: The model answer for this exercise is too complicated: $66 {2\over 3} = 66 + {2\over 3} = {200 \over 3}$. So $66 {2\over 3} \% = {200 \over 300} = {2 \over 3}$. Similarly, $16 {2\over 3} = {50 \over 3}$ and $16 {2\over 3}\% = {50 \over 300} = {1 \over 6}$.
A: The recurring decimal $0.\overline{a_1\ldots a_n}$ is equal to
$$\frac{a_1\cdots a_n}{10^n-1}.$$
E.g., $x=0.\overline{285} = 0.285285285\cdots$, then
$$x = \frac{285}{10^3-1} = \frac{285}{999}.$$
That is, you get the periodic portion divided by a number that consists of as many $9$s as the length of the periodic portion.
There are many ways of seeing this; one is using geometric series. Another is to use some manipulations: if
$$x = 0.\overline{a_1\ldots a_n}$$
then
$$10^nx = a_1\ldots a_n . \overline{a_1\cdots a_n}$$
so
$$(10^n-1)x = 10^n x - x  = a_1\cdots a_n.$$
The first "model solution" is using this: since $x = 0.\overline{6}$, then $x = \frac{6}{9}$ (the period has length $1$, so you get a single $9$ in the denominator.
When the periodic decimal does not start right after the decimal point, you need to shift it a bit first. So for example, if you had
$$ x = 0.1\overline{285} = 0.1285285285\ldots,$$
then first we take $10 x = 1.\overline{285}$, then proceed as before:
$$\begin{align*}
10^3(10 x) &= 1285.\overline{285}\\
 10x &= 1.\overline{285}\\
10x(10^3-1) &= 1284\\
x(9990)&= 1284\\
x &= \frac{1284}{9990}.
\end{align*}$$
The second model solution uses this method.
Added. For the series method, in case anyone is interested, suppose that $x$ is of the form $x=0.\overline{a_1\cdots a_n}$. This means, explicitly, that
$$ x = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{(10^n)^k} = \sum_{k=1}^{\infty}\frac{a_1\cdots a_n}{10^{nk}} = a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}}.$$
This is a geometric series, with initial term $\frac{1}{10^{n}}$ and common ratio $\frac{1}{10^n}$, so it converges. A geometric series with initial term $a$ and common ratio $r$, $|r|\lt 1$, converges to
$$\frac{a}{1 - r},$$
so we have
$$\begin{align*}
x &= a_1\cdots a_n\sum_{k=1}^{\infty}\frac{1}{10^{nk}} \\
&= a_1\cdots a_n\left(\frac{\frac{1}{10^n}}{1 - \frac{1}{10^n}} \right)\\
&= a_1\cdots a_n\left(\frac{\quad\frac{1}{10^n}\quad}{\quad\frac{10^n-1}{10^n}\quad}\right)\\
&= a_1\cdots a_n\left(\frac{1}{10^n-1}\right) = \frac{a_1\cdots a_n}{10^n-1}\\
&= \frac{a_1\cdots a_n}{\underbrace{9\cdots 9}_{n\text{ digits}}}.
\end{align*}$$
And similarly if you have to "shift" the decimal before you get to the period; you simply add enough $0$s to the $9$s in the denominator to account for the shift.
A: It happens to be well known that a number over 9 (other than 0 or 9) has a repeating decimal. That is: $\dfrac{1}{9} = 0.11 \bar{1}$, $\dfrac{4}{9} = 0.44 \bar{4} $, and so on. Why is this true?
It comes from the fact that if we have the equation $10x = 1.\bar{1}$, then we have $x = 0.\bar{1}$ by division. Subtracting, we get that $9x = 1$, or that $x = \dfrac{1}{9}$.
As happens - I typed this at the same time as Ross.
