Why does dividing a number with $n$ digits by $n$ $9$'s lead to repeated decimals?

Question

For example, $\frac{1563}{9999} = 0.\overline{1563}$.

Why does that make sense from the way the number system works?

I can vaguely see that since the number $b$ with $n$ $9$'s is always greater than the dividend $a$ of $n$ digits (except in the corner case where they are the same), you will always get $0$ for the units place, after which in the long division process you will be able to divide $a \cdot 10^n$ by $10^n$, $a$ times, which implies you'll be able to divide $a \cdot 10^n$ by $b$, $a$ times with a remainder of $a$, thereby perpetuating this process over and over.

However, I feel this isn't a very precise way to think about it...

Answer 1

$$ \dfrac{a}{10^n} + \dfrac{a}{10^{2n}} + \dfrac{a}{10^{3n}} + \ldots = \dfrac{a}{10^n - 1} $$ (by the formula for a geometric series)

Answer 2

This is because the example can be rewritten as \begin{align*}\frac{1563}{9999}&=0.1563\cdot\frac 1{1-10^{-4}}=0.1563(1+0.0001+0.00000001+\dots)\\ &=0.1563+0.00001563+0.000000001563+\dots \end{align*}

Answer 3

Another way to think about it is $$ \frac{1563}{9999} - \frac{1563}{10000} = \frac{1563}{9999 \cdot 10000} $$ which inductively repeats itself. It is easily generalised to any lengths and any bases.

Answer 4

This happens because that number $10^n-1=$bunch of $9$'s (for $n$ a natural number) is not divisible by only $2$ and $5$. Any denominator with prime factors other than $2$ and $5$ will have a repeating decimal (it maybe fairly long).