Did I discover something new? How do I prove it?

Question

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:

Answer 1

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.

Answer 2

$$\frac{10}{81}=0.1234567901234612345679012346\cdots=0.123456789+0.000000000123456789+\cdots$$

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.

Answer 3

The basic operation you are doing to observe the pattern is to consider $3n\cdot 123456789 + 3n = n(3\cdot 123456789+3)$. Now $3\cdot 123456789+3 = 370 + 370\cdot 10^3 + 370\cdot 10^6$. There's nothing special about this number, so let's instead fix some positive integers $k$ and $N$ with $k < 10^N$ and $r$ and consider $$y = \sum_{s=0}^r k10^{sN}.$$ In your example, $k = 370$, $N = 3$, and $r = 2$, so $y = 3\cdot 123456789+3$. We want to show that for $n \geq 1$ the quantity $ny +$ some leading digits of $ny$ has a pattern of $N$ digits repeating $r+1$ times. It turns out that this will be true provided that $ny \leq 10^{(2r+1)N}-3$. In your case, this means that this will be true so long as $n \leq 10^{6.4}$, so it's no wonder you didn't find a counterexample.

To be more explicit, in your example, $n = 4$, and $ny = 481481480$, and the leading digit we need to add is a $4$, and we repeat the sequence $481$ a total of $3$ times. If $n = 271$, then $ny = 100370370270$ and we need to add on the leading digits $100$ to get $ny + 100 = 100370370370$, and repeat the sequence $370$ a total of $3$ times.

Let's take a small detour before tackling the problem. Let $m$ be any integer with decimal expansion $d_1d_2\ldots d_M$ and consider the decimal expansion of $\frac{m}{10^M-1} = \frac{d_1\ldots d_M}{9\ldots 9}$. Recall that this is just $0.\overline{d_1\ldots d_M}$. Indeed, we can express this decimal as $$(d_1\ldots d_M)(10^{-M}) + (d_1\ldots d_M)(10^{-2M}) + (d_1\ldots d_M)(10^{-3M}) + \cdots = \sum_{s = 1}^{\infty}m(10)^{-sM} = \frac{m 10^M}{1-10^{-M}} = \frac{m}{10^M-1}.$$

Now back to the original problem. By definition $$ny = \sum_{s=0}^r (nk)10^{sN}.$$ Then, with $M = (r+1)N$ $$\sum_{t=0}^{\infty}ny 10^{-M} = nk\sum_{t=0}^{\infty}\sum_{s=0}^r 10^{sN}10^{-M} = nk \frac{10^{(r+1)N}-1}{(1-10^{-(r+1)N})(10^N-1)} = 10^{(r+1)N}\frac{nk}{10^N-1}.$$

Let us write $nk = p\cdot (10^{N-1}-1) + q$, where $q \leq 10^{N-1}-1$ (it's convenient to allow $q = 10^{N-1}-1$ if possible, so that $q \neq 0$ if $n \neq 0$). Thus $$10^{(r+1)N}\frac{nk}{10^N-1} = 10^{(r+1)N}p + 10^{(r+1)N}\frac{q}{10^N-1}.$$ The first term is an integer with $N(r+1)$ tailing zeroes in its decimal expansion, say $e_1\ldots e_K0\ldots 0$. We we know from the detour that $\frac{q}{10^N-1}$ is a repeating decimal less than $1$, and each repeated block has length $N$; suppose it is $0.\overline{d_1\ldots d_N}$. Then the second term is $(d_1\ldots d_N)\cdots(d_1\ldots d_N).\overline{d_1\ldots d_N}$, where the block before the decimal point repeats $r+1$ times.

It follows that $$\sum_{t=0}^{\infty}ny 10^{-M} = 10^{-M}10^{(r+1)N}\frac{nk}{10^N-1} = e_1\ldots e_K (d_1\ldots d_N)\cdots(d_1\ldots d_N).\overline{d_1\ldots d_N}.$$

However, notice that $\sum_{t=2}^{\infty} ny 10^{-M} = 10^{-2M}\frac{ny}{1-10^{-2M}} = \frac{ny}{10^{2M}-1} < 10^{-N}$ since $ny \leq 10^{(2r+1)N}-3$. Thus $$\sum_{t=0}^{1}ny 10^{-M} = ny + ny/10^M = e_1\ldots e_K (d_1\ldots d_N)\cdots(d_1\ldots d_N).\overline{d_1\ldots d_N} - x$$, where $x$ is some positive number less than $10^{-N}$. In particular, so long as $d_1\cdots d_N$ is not identically $0$, $0 \leq 0.\overline{d_1\ldots d_N} - x < 1$ and so $$ e_1\ldots e_K (d_1\ldots d_N)\cdots(d_1\ldots d_N) \leq ny + ny/10^M < e_1\ldots e_K (d_1\ldots d_N)\cdots(d_1\ldots d_N) + 1,$$ and so $ny$ + some leading digits has the repeating pattern $d_1\ldots d_N$ a total of $r+1$ times.

However it's not possible for $d_1\cdots d_N$ to be identically $0$ unless $n = 0$, since this would necessitate $q = 0$.

This proof also predicts the periodicity of the pattern: just take $nk \pmod {10^N-1}$, with the usual residue class, except if $nk \equiv 0 \pmod {10^N-1}$, in which case the pattern is $99\ldots 9$. This explains the periodicity of your patterns, as well. In your example, $k = 370$, and $370\cdot 27$ is the lcm of $370$ and $999 = 10^3 - 1$, and so the pattern repeats each every 27 integers, except with $999$ instead of $000$. The occurrence of 27, which is about $10e$, appears to be purely coincidental.