Many years ago, I noticed that $987654321/123456789 = 8.0000000729\ldots$.
I sent it in to Martin Gardner at Scientific American and he published it in his column!!!
My life has gone downhill since then:)
My questions are:
Why is this so?
What happens beyond the "$729$"?
What happens in bases other than $10$?
On
$$729=9^3$$ $$66339=9^3\cdot 91$$ $$6036849=9^3\cdot 91^2$$ $$...$$ $$987654321/123456789=8+9^3\cdot 10^{-10}\cdot\displaystyle\sum_{n=0}^{\infty}(91\cdot 10^{-10})^n$$
On
Let $$S_n(a)=1 +2a+\ldots +na^{n-1}=\frac{na^{n+1}-(n+1)a^n+1}{(a-1)^2},$$ $$T_n(a)=a^{n-1}+2a^{n-2}+\ldots +n=a^{n-1}S_n(a^{-1}).$$
Then $$ \frac{S_n(a)}{T_n(a)}=\frac{na^{n+1}-(n+1)a^n+1}{a^{n+1}-(n+1)a+n}.$$ For $a=10,n=9$ we have $$ \frac{S_n(a)}{T_n(a)}\approx\frac{8\cdot 10^{10}+1}{10^{10}}. $$
On
Just to add to the excellent answers above, some examples:
${987654321\,/\,123456789}\approx 8.00000007290000066339$
${{87654321}_9\,/\,{12345678}_9}\approx {7.000000628000056238}_9$
${{7654321}_8\,/\,{1234567}_8}\approx {6.0000052700046137}_8$
${{654321}_7\,/\,{123456}_7}\approx {5.00004260036036}_7$
${{\mathrm{fedcba987654321}}_{16}\,/\,{\mathrm{123456789abcdef}}_{16}}\approx {\mathrm{e.0000000000000d2f00000000000c693f}}_{16}$
On
$98765432 / 12345679 = 8$, exactly. You can see how the pattern works by multiplying out $12345679 * 8$ starting at either end.
This explains why your fraction is close to an integer. If you think the $729$ is interesting (I don't), it can be explained by some of the other answers here.
What can we say about the fact that $12345679 * 8 = 98765432$? I have been aware of this 'factlet' for about 20 years, and remember it being used to 'demonstrate' calculators (which often had 8 digit displays back in the day).
I just recently realised that:
$$ \frac{1}{81} = \left(\frac{1}{9}\right)^2 = \left(\sum_{k=1}^{\infty}\frac{1}{10^k}\right)^2 = \sum_{k=1}^{\infty} \sum_{m=1}^{k-1} \frac{1}{10^m} \frac{1}{10^{k-m}} = \sum_{k=1}^{\infty} \frac{k-1}{10^k} $$
In other words, while $\frac{1}{9} = 0.1111111\ldots$ $$ \frac{1}{81} = 0.01 + 0.002 + 0.0003 + 0.0004 + 0.00005 \ldots $$
It is pretty easy to see that this infinite sum is going to converge to something starting $0.012345\ldots$. If you keep on adding, or work out $\frac{1}{81}$ by division, you get $$ 0.012345679012345679012345679\ldots $$ When you get to the point where you add $\frac{10}{10^{11}}$, the first carry happens, which leads to the 9 where you might expect an 8. After that every addition carries and the decimal expansion repeats every 9 digits (not every ten - because the amount we carry keeps on getting bigger and bigger).
Now, $\frac{8}{81} = \frac{9}{81} - \frac{1}{81}$, or $$ \frac{8}{81} = 0.11111111\ldots - 0.012345679012345\ldots $$ Think of each '1' digit in $0.111\ldots$ as being a '10' in the next column. This means that we can work out $\frac{8}{81}$ as the "10's complement" of $\frac{1}{81}$, since we are subtracting a digit between $1$ and $9$ from $10$, to get another single digit which appears in the same place. So $\frac{8}{81}$ starts $0.098765\ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2.
So $$ \frac{8}{81} = 0.098765432098765432098765\ldots $$
and therefore $$ 0.0123456790123456790\ldots * 8 = 0.0987654320987654320\ldots $$ and clearly this gets you that $$ 12345679 * 8 = 98765432 $$
In base $n$ the numerator is $$p = n^{n-1} - \frac{n^{n-1}-1}{(n-1)^2}$$ and the denominator is $$q = \frac{n(n^{n-1}-1)}{(n-1)^2}-1.$$
Note that $p = (n-2)q + n-1$ and for the quotient we get
\begin{align} \frac{p}{q} &= n-2 + \frac{(n-1)^3}{n^n} \frac{1}{1 - \frac{n^2-n+1}{n^n}} \\ &= n-2 + \frac{(n-1)^3}{n^n} \sum_{k=0}^{\infty} \left(\frac{n^2-n+1}{n^n}\right)^k. \end{align}
Indeed for $n=10$ this is
$$\frac{987654321}{123456789} = 8 + \frac{729}{10^{10}}\sum_{k=0}^{\infty}\left(\frac{91}{10^{10}}\right)^k $$