Simple Repetition

Let $x=0.1212...$ . Then $100x=12.1212...$ . Now $100x-x=12.1212...-0.1212...=12$ which means that $99x=12$ . Hence $x=\frac{12}{99}=\frac{4}{33}$ .

$\begin{aligned} \text{We note that } \space 0.1212... & = 12 \times 0.010101... \\ & = 12 \times (0.01 + 0.0001 + 0.000001 + ... ) \\ & = 12 \times \sum_{n=1}^\infty \left(\frac{1}{100}\right)^n \\ & = 12 \times \left(\frac{1}{100}\right)\left(\frac{1}{1-\frac{1}{100}}\right) \\ & = \frac{12}{99} = \boxed{\dfrac{4}{33}} \end{aligned}$

In general:

$\begin{array}{lll} 0.\dot{1} & = 0.111... & = \dfrac{1}{9} \\ 0.\dot{1}\dot{2} & = 0.121212... & = \dfrac{12}{99} \\ 0.\dot{1}\dot{2}\dot{3} & = 0.123123123... & = \dfrac{123}{999} \\ ... & \quad ... & \quad ... \\ \end{array}$

$\Rightarrow n$ repeated digits with $n$ digits of $9$ in the denominator.

Let x=0.121212.... so,100x=12.121212.... now,100x-x=12.121212...-0.121212... so,99x=12 hence,x=12/99 or,x=4/33 Therefore,x=4/33

Let $x=0.\overline{12}$ ,so $100x=12.\overline{12} \implies 100x-x=12.\overline{12}-0.\overline{12}=12 \implies 99x=12 \implies x=\frac{12}{99}=\boxed{\large{\frac{4}{33}}}$

The only solutions that evaluate to more than 0.1 are 3/25 and 4/33. 3/25 is not a repeating decimal (it's 12/100 or 0.12). By process of elimination, the solution must be 4/33.

I kind of found this in a book, actually. I was younger, I forgot the title.

1 ÷ 8.3 = 0.1204819277 and 1 ÷ 8.2 = 0.1219512195. So the correct answer should be somwhere in between 8.2 an 8.3. Therefore lets try 1 ÷ 8.25 and the answer is = 0.1212121212 bulls eyes :-) ! So 8.25 x 4 = 33 So the answer is 4÷33.