Did I discover something new? How do I prove it? I saw recently a post on a math page about the following interesting approximation:

Which made me play around with the number at the denominator, 123456789, and I discovered that if you multiply it by a multiple of 3 you are going to get a interesting number, which I believe an example might be worth in order to comprehend.
Let's multiply 123456789 by 12, a multiple of 3:

Take the result and add 12, the number that you multiplied 123456789 by:

See that division of the number in sequences of 3 digits from right to left? After 3 sequences of 3 digits, take what comes after (In this case, one) and add it to the previous result:

That same 3 sequences of 3 digits are going to be exactly the same after the process, which in this example was 481, and that's the problem: That same 3 sequences of 3 digits are going to be exactly the same after the process, which in this example was 481, and that's the problem: how do you prove that this is always going to happen (if it always does)? I've already tried to check the first 500 cases and I didn't see an exception.
Also, if we say that the number (r) we get after multiplying 123456789 by a multiple of 3 is:

And if we look at it's table of values, that means, a table of the corresponding r for a given n, we see that after 27 units, the 3 digit sequence that repeats is going to reset to the first one, when n=1, which is 370 (note that x=n and the huge equation is equal to the respective 3 digit sequence):


So, the graph of r(n) is going to be:

 A: $$\frac{10}{81}=0.1234567901234612345679012346\cdots=0.123456789+0.000000000123456789+\cdots$$
This fractional number has the period $9$. When you multiply it by $3$, you get a number with period $3$,
$$\frac{10}{27}=0.370370370370370370370370\cdots=0.370+0.000370+\cdots$$
and again by $3$,
$$\frac{10}{9}=0.111\cdots=0.1+0.01+0.001+\cdots$$
As you can imagine,
$$\frac{10}{243}=0.0411522633744855967078189300411522633744855967078189\cdots$$ has period $27$.
If you multiply these numbers by an integer which is not a multiple of $3$, the period remains unchanged. Sorry, there is no deep math here.
A: We have, exactly:
$\dfrac{0.\overline{098765432}}{0.\overline{012345679}}=8$
The numerator is $8/81$, the denominator is $1/81$.  Basically the quoted fraction follows this pattern except in the last couple digits if the numerator and denominator.
Similar constructions exist in other bases, the quotient being close to the base minus $2$.  For instance in base $5$:
$\dfrac{4321}{1234}\approx \dfrac{0.\overline{0432}}{0.\overline{0124}}=3$
In base $5$ we actually have $1234×3=4312$, so again we're off only in the last couple "digits".  However, with fewer "digits" overall the near-integer approximation is not as good.  It becomes more accurate in higher bases.
