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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1This doesn't quite answer your question, but it's very similar... – Micah Dec 28 '14 at 03:53
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5Because $\frac{98765432}{10^9-1}=\frac{8}{81}$ – Math.StackExchange Dec 28 '14 at 03:56
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Really? I need to calculate that. – ReliableMathBoy Dec 28 '14 at 03:59
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2Yeah, you're right, but remember to answer this question in the answer section. – ReliableMathBoy Dec 28 '14 at 04:00
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Why is it missing the $1$? – ReliableMathBoy Dec 28 '14 at 04:06
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Look at 1/81, that's also interesting. – Akiva Weinberger Dec 31 '14 at 15:38
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It's true. You get every single one-digit number except for one of them: $0.012345679012345679.... It's missing the eight because of some addition explained in the "linked" area. – ReliableMathBoy Jan 01 '15 at 17:59
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Sorry, I meant to make it $0.012345679012345679...$. – ReliableMathBoy Jan 01 '15 at 18:02
2 Answers
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The link that @Micah posted in the comment to your post is a good explanation.
Maybe you can also see it by the following calculations
$$x = 0.\overline{098765432}$$
$$10^9 x = 98765432.\overline{098765432}$$
$$(10^9 - 1)x = 98765432$$
$$x = 8/81$$
We could actually find the number $y = 0.\overline{0987654321}$ by similar logic.
$$10^{10} y = 987654321.\overline{0987654321}$$
$$(10^{10} - 1) y = 987654321$$
$$y = \frac{109739369}{1111111111}$$
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 $y$.
Chinmay Nirkhe
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More generally, in base $b$ (any integer $>1$)
$$ \dfrac{b-1}{b^2} + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} = \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}$$ so that $$ \eqalign{ \dfrac{b-1}{b^2}& + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} + \dfrac{0}{b^{b}} + \dfrac{b-1}{b^{b+1}} + \dfrac{b-2}{b^{b+3}} + \ldots = \sum_{j=0}^\infty b^{-j(b-1)} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}\cr &= \dfrac{b^{b-1}}{b^{b-1}-1} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}= \dfrac{(b-2)}{(b-1)^2}}$$
What you have is the case $b=10$.
Robert Israel
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