# Why do I get 0.098765432098765432... when I divide 8 by 81? I got this remarkable thing when

Why do I get $0.098765432098765432...$ when I divide 8 by 81?
I got this remarkable thing when I divided 16 by 162, or, in a simplified version, 8 by 81. It's $0.098765432098765432\cdots$ , or more commonly known as $0.\overline{098765432}$ with all the one-digit numbers going backwards...except for 1. Yeah, it's missing the 1. One, how do I get this remarkable outcome and two, why is it missing the 1?
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Leah Conley
Maybe you can also see it by the following calculations
$x=0.\overline{098765432}$
${10}^{9}x=98765432.\overline{098765432}$
$\left({10}^{9}-1\right)x=98765432$
$x=8/81$
We could actually find the number $y=0.\overline{0987654321}$ by similar logic.
${10}^{10}y=987654321.\overline{0987654321}$
$\left({10}^{10}-1\right)y=987654321$
$y=\frac{109739369}{1111111111}$
So maybe the reason why you're number is cool is that it has a easy fractional representation and a decimal representation with interesting properties. Any interesting repeating decimal representation is of course a fraction and can be calculated, but it may not have an easy fractional representation as shown with y.

hughy46u
More generally, in base b (any integer $>1$)
$\frac{b-1}{{b}^{2}}+\frac{b-2}{{b}^{3}}+\dots +\frac{2}{{b}^{b-1}}=\frac{\left(b-2\right)\left({b}^{b-1}-1\right)}{{b}^{b-1}\left(b-1{\right)}^{2}}$
so that
$\begin{array}{rl}\frac{b-1}{{b}^{2}}& +\frac{b-2}{{b}^{3}}+\dots +\frac{2}{{b}^{b-1}}+\frac{0}{{b}^{b}}+\frac{b-1}{{b}^{b+1}}+\frac{b-2}{{b}^{b+3}}+\dots =\sum _{j=0}^{\mathrm{\infty }}{b}^{-j\left(b-1\right)}\frac{\left(b-2\right)\left({b}^{b-1}-1\right)}{{b}^{b-1}\left(b-1{\right)}^{2}}\\ & =\frac{{b}^{b-1}}{{b}^{b-1}-1}\frac{\left(b-2\right)\left({b}^{b-1}-1\right)}{{b}^{b-1}\left(b-1{\right)}^{2}}=\frac{\left(b-2\right)}{\left(b-1{\right)}^{2}}\end{array}$
What you have is the case $b=10$