Thanks so much for your responses and I've been still staring at

these things.

So the summary is:

[P, Q, are primes other than 3; R is the repeating part of the

repetend, n is the number of repeating digits]

statement 1: P/Q always yields some R

statement 2: n <= Q

statement 3: R mod 9 = 0

and I might add a couple more trivial ones:

R is always odd (I.E not 18, 36 etc..)

Also R mod P = 0

so really R mod 9P = 0

It seems that the numerator prime P is irrelevant to this discussion

acnd could be factored out for simplicity,

1/Q yields a repetend "r" where r mod 9 = 0.

So this means each prime has a corrosponding r/9 number.

Question: Is it necessarily unique?; I would think so, but I've never

been too good at thinking through all the logical consequences of

that.

-Shawnbob

--- In

primenumbers@yahoogroups.com, Alan Eliasen <eliasen@m...>

wrote:

> patience_and_fortitude wrote:

> > If you divide one prime by another, it generally results in a

> > repeating decimal of some length.

>

> It might be helpful to clarify that if the denominator has

factors of

> anything but 2 and/or 5, it will *always* produce a repeating

decimal.

>

> If the denominator has only factors of 2 and 5, or, more

specifically

> 2^n and 5^m, the decimal part will terminate after at most max(n,m)

digits.

>

> The maximum length of the part that repeats can be no bigger

than the

> denominator. That is, if the denominator is, say, 17, the decimal

will

> repeat after 17 digits or less. You can see why this happens by

working

> some samples out with long division.

>

> --

> Alan Eliasen | "It is not enough to do your best;

> eliasen@m... | you must know what to do and THEN

> http://futureboy.homeip.net/ | do your best." -- W. Edwards

Deming