Why does this approximation for 8 work? I previously saw that peculiarly $\frac{987654321}{123456789}\approx 8$. I was wondering if there was any significance to it i.e. if there is any way to derive this approximation (aside from long division).
I already shared my solution. I would love to see other interesting methods (if any).
 A: More generally, in base $b$, the number with digits decreasing from $b-1$ to $1$ is
$$ N = \sum_{i=0}^{b-2} (i+1) b^i = \frac{b-2}{(b-1)^2} b^{b} + \frac{1}{(b-1)^2}$$
while the number with digits increasing from $1$ to $b-1$ is
$$D = \sum_{i=0}^{b-2} (b-1-i) b^i =  \frac{b^b}{(b-1)^2} - \frac{b^2-b+1}{(b-1)^2}$$
For large $b$, the dominant terms are those with $b^b$, so
$$ \frac{N}{D} \sim b-2 $$
A: We can express $987654321$ as approximately $\sum_{k=-9}^\infty -k\cdot 10^{-k-1}$ and $123456789$ as approximately $\sum_{k=1}^\infty k\cdot 10^{9-k}$
If we take
$$f(x)=\sum_{k=-9}^\infty x^k=\frac{x^{-9}}{1-x}$$
then
$$f'(x)=\sum_{k=-9}^\infty kx^{k-1}=\frac{-9x^{-10}+10x^{-9}}{(1-x)^2}$$
$$\implies f'(x)=\sum_{k=-9}^\infty kx^{k-1}=\frac{-9x^{-10}+10x^{-9}}{(1-x)^2}$$
$$\implies x^2f'(x)=\sum_{k=-9}^\infty kx^{k+1}=\frac{-9x^{-8}+10x^{-7}}{(1-x)^2}$$
$$\implies -10^2\cdot f'\left(\frac{1}{10}\right)=\sum_{k=-9}^\infty -k\cdot 10^{-k-1}=\frac{8\cdot 10^{10}}{81}$$
Similarly, if we take
$$g(x)=\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$
then
$$g'(x)=\sum_{k=1}^\infty k\cdot x^{k-1}=\frac{1}{(1-x)^2}$$
$$\implies x^{-8}g'(x)=\sum_{k=1}^\infty k\cdot x^{k-9}=\frac{x^{-9}}{(1-x)^2}$$
$$\implies 10^8\cdot g'\left(\frac{1}{10}\right)=\sum_{k=1}^\infty k\cdot 10^{9-k}=\frac{10^{10}}{81}$$
Hence, from our approximations for $987654321$ and $123456789$, it follows that $\frac{987654321}{123456789}\approx 8$
A: We have that in base $8$

*

*$987654321=7267464261_8$

*$123456789=726746425_8$
then
$$987654321-8\cdot 123456789=7267464261_8-10_8\cdot 726746425_8=11_8=9$$
