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

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...

• Hint:9999 = 10000 -1. Multiply both sides by 10000-1. You gey 1563 = 1563.1563.... - .1563 = 1563. Mar 14, 2016 at 1:48
• Notice that the number does not necessarily have to have the same number of digits as the string of $9$'s. Mar 14, 2016 at 1:49
• You might find it interesting to try and gain an understand of how long those repeating decimals will be based on how the denominator factors. Mar 14, 2016 at 1:50

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