# Talk:81 (number)

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## Contents

I don't think the link to -20 is what this article wants. Check to see if it is correct. Georgia guy 00:51, 31 Jan 2005 (UTC)

I just heard an unconfirmed report that a new number has been discovered between 80 and 81. If true, this could have major implications for number 81. —Preceding unsigned comment added by Kackermann (talkcontribs) 14:31, 21 October 2007 (UTC)

## Specific and general

That ${\displaystyle {\frac {1}{81}}}$ exhibits the sequence of digits except 8 is just the base-10 case of the general rule that ${\displaystyle {\frac {1}{(n-1)^{2}}}}$ for any base n gives the sequences of digits except n−2. Cf. ${\displaystyle {\frac {1}{121}}}$ in base 12, ${\displaystyle {\frac {1}{225}}}$ in base 16 (0.0123456789ABCDF…) to name just two. --109.67.200.119 (talk) 16:13, 10 June 2010 (UTC)

I think you're right. Perhaps we should say that in the article. In fact, I will. — Arthur Rubin (talk) 17:09, 10 June 2010 (UTC)
Thanks. It's of high probability I'm right, because I tested this on a great number of bases using bc, and they all bear it out. However, as with trying out a million triangles of two equal edges and one equal angle and finding out they overlap, that only makes for high probability, not mathematical proof. I don't know a proof. --109.67.200.119 (talk) 17:35, 10 June 2010 (UTC)
Let me see....(I don't know how to do alignments in TeX.)
${\displaystyle 123\ldots (b-4)(b-3)(b-1)_{b}=1+\sum _{k=0}^{b-2}b^{k}(b-2-k)}$
${\displaystyle =1+\sum _{l=0}^{b-3}\sum _{k=0}^{l}b^{k}}$
${\displaystyle =1+\sum _{l=0}^{b-3}{\frac {b^{l+1}-1}{b-1}}}$
${\displaystyle =1-{\frac {b-2}{b-1}}+{\frac {1}{b-1}}\sum _{l=0}^{b-3}b^{l+1}}$
${\displaystyle ={\frac {1}{b-1}}+{\frac {1}{b-1}}\sum _{l=0}^{b-3}b^{l+1}}$
${\displaystyle ={\frac {1}{b-1}}\sum _{l=0}^{b-2}b^{l}}$
${\displaystyle ={\frac {b^{b-1}-1}{(b-1)^{2}}}.}$

(Informative) Also applies to signed-digit bases. ${\displaystyle {\frac {1}{676}}}$ (676 = square of 26) in balanced base 27 (which stands for three balanced ternary digits much like hexadecimal stands for four binary digits): ${\displaystyle 0.0123456789ABD{\overline {DCBA987654321}}0123...}$ (the vinculum marks the negative digits). --79.178.202.78 (talk) 17:06, 26 July 2010 (UTC)

## It is also the only number…

Where the sum of its digits equal its square root. Is anyone going to revert this if I make an article change? —Preceding unsigned comment added by Da5id403 (talkcontribs) 23:48, 12 May 2011 (UTC)

## Hell's Angels

81 is also the number for the Hells Angels MC. 8=H 1=A — Preceding unsigned comment added by 71.231.251.27 (talk) 16:37, 8 September 2014 (UTC)