Not just a single number, but a kind of number

To get a repeating decimal of any pattern you want of finite length, you can multiply your digit $x$ who has $n$ digits by $0.\overline{\underbrace{00...0}_{n-1}1}$ .

This number has enough $0$ 's to hold your number, and multiplies it by 1, then just keeps concocting them. Now we need to find this number as a fraction.

Lets look at the fraction $\dfrac{1}{9}$ . As a decimal it is $0.\overline{1111}$ . This is a good baseline, as this $\dfrac{x}{9}$ is the 1 digit repeating number fraction.

Lets look at the fraction $\dfrac{1}{11}$ . As a decimal it is $0.\overline{0909}$ . We now have a 2 digit repeating decimal, but we need $1$ 's, not $9$ 's. So lets multiply by $\dfrac{1}{9}$ . $\dfrac{1}{11}\times\dfrac{1}{9}=0.\overline{0909}\div9=0.\overline{0101}=\dfrac{1}{99}$ . We can see multiplying this decimal by any $2$ digit number would give us the $2$ digit repeating decimal. So $\dfrac{x}{99}$ is the 2 digit repeating fraction. (The pattern is forming).

This pattern does indeed repeat, to get $0.\overline{009009}$ you need the fraction $\dfrac{1}{111}$ . Repeat the steps as above and you get the 3 digit repeat fraction of $\dfrac{1}{999}$ .

Recapping:

$1$ digit $=\dfrac{1}{9}=\dfrac{1}{10^1 -1}$

$2$ digits $=\dfrac{1}{99}=\dfrac{1}{10^2 -1}$

$3$ digits $=\dfrac{1}{9}=\dfrac{1}{10^3 -1}$

$\dots$

$n$ digits $=\dfrac{1}{9}=\dfrac{1}{10^n -1}$

$n+1$ digits $=\dfrac{1}{9}=\dfrac{1}{10^{n+1} -1}$ . Our number has $n+1$ digits, therfore $\chi=10^{n+1} -1$

P.S. I already had prior knowledge about how repeating decimals worked, so I kind of just described the pattern. Somebody please show the work of deriving $\dfrac{1}{9} and \dfrac{1}{\underbrace{111\dots111}_{x}}$ as the key fractions needed

$\begin{aligned} \dfrac{N}{\chi} &= 0. \ \overline{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0} \\ \implies 10^{n+1} \times \dfrac{N}{\chi} &= \overline{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0} \ . \ \overline{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0} ~~~~~~~~~~~~~~~~~~~~~~ \small\color{#3D99F6}{\text{As number of recurring digits is} \ n+1} \end{aligned} \\ \therefore \left( 10^{n+1} - 1 \right) \dfrac{N}{\chi} = \overline{a_n a_{n-1} a_{n-2} ... a_2 a_1 a_0} \\ \implies \left( 10^{n+1} - 1 \right) \dfrac{N}{\chi} = N \\ \implies \left( 10^{n+1} - 1 \right) \dfrac{1}{\chi} = 1 \\ \implies \chi = \boxed{10^{n+1} - 1}$