First of all, note that 1/998001 = (1/999)^2. 1/999 = 10^-3 + 10^-6 + 10^-9 + ... Hence, 1/998001 = 10^-6 + 2*10^-9 + 3*10^-12 + ... = 0.000001002003004005006007008 ... 99699799899X001002003004005 ... (where X represents ten) = 0.000001002003004005006007008 ... 996997999000001002003004005 ... 998 should be in there, in the substring "... 798799800801 ..." Let IO(1000) = 1/998001 -- the 'imperfect odometer' in base 1000. It is clear that IO(n) = 1/(n-1)^2. However, what if we want a *perfect* odometer in base n (which doesn't omit 'n-2' from its n-ary expansion)? For this, we'll need a rational with a much larger denominator: 0.01234567890123456789012345 ... = 123456789/999999999 It is easy to find a closed form for '9999999999' generalised to any base, namely n^n - 1. '123456789' is a little trickier, but it is obvious that: 123456789 * 10 = 1234567890 = 123456789 + 1111111101 Therefore 123456789 * 9 = 1111111101, 123456789 * 81 = 9999999909 = 10^10 - 10^2 + 10 - 1. The generalisation of 123456789 is (n^n - n^2 + n - 1)/(n-1)^2. So, our 'perfect odometer' PO(n) = (n^n - n^2 + n - 1)/((n^n-1)(n-1)^2). Sincerely, Adam P. Goucher ----- Original Message ----- From: "Ray Tayek" <rtayek@ca.rr.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, January 27, 2012 11:31 AM Subject: Re: [math-fun] Fun with math: Dividing one by 998001 yields a surprising result

At 03:05 AM 1/27/2012, you wrote:

...

http://www.iheartchaos.com/post/16393143676/fun-with-math-dividing-one-by-99...

can't seem to find 998 in there.

thanks

--- co-chair http://ocjug.org/

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