<s>[INST] 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?
 [/INST] 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.
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