Why does $987,654,321$ divided by $123,456,789 = 8$?

Question

Why does $987,654,321$ divided by $123,456,789 = 8$?

Is it a coincidence or is there a special reason?

Note: The numbers are a mirror of each other

Answer 1

$987654321 - 8 \cdot 123456789 = 9$ or $$ \frac{987654321}{123456789} = \; 8 \; + \; \frac{1}{13717421} = \; 8 \; + \; \frac{729}{9999999909} \approx 8.0000000729$$ note the $0$ in the tens place in 9999999909

Answer 2

$$123456789*8=9876543{\color{red}1\color{red}2}$$

The number changes the last two digits, so that division can't be correct.

Answer 3

I think whatever calculator you did the calculation with made a rounding error. The quotient of your two numbers in question is close to 8, but it is not 8 exactly.

Answer 4

$$123456789*9=1111111101\\ 123456789+987654321=1111111110$$

Answer 5

Just to generalize:

$$\sum_{i=1}^{b-1} ix^{i-1} = \left(\frac{1-x^b}{1-x}\right)'=\frac{(b-1)x^b-bx^{b-1}+1}{(x-1)^2}$$

Letting $b=x=10$ you get:

$$987654321=\frac{9\cdot 10^{10}-10\cdot 10^9 + 1}{9^2}=\frac{8\cdot 10^{10}+1}{9^2}$$

Then $123456789-111111111 = 999999999-987654321$ so $$\begin{align}123456789&=10\frac{10^{9}-1}{9}-\frac{8\cdot 10^{10}+1}{9^2}\\ &=\frac{10^{10}+1-90}{9^2}\end{align}$$

So $$987654321-8\cdot 123456789 = \frac{1+8\cdot (10^2-9)}{9^2}=9$$

More generally, in base $b$:

$$A=((b-1)(b-2)\cdots 1)_b = \frac{(b-2)b^b+1}{(b-1)^2}$$

And:

$$\begin{align}B&=(123\cdots (b-1))_b \\ &= \frac{b^b-b}{b-1} - \frac{(b-2)b^b+1}{(b-1)^2}\\ &=\frac{b^b-(b^2-b+1)}{(b-1)^2} \end{align}$$

Then $$A-(b-2)B= \frac{1+(b^2-b+1)(b-2)}{(b-1)^2}=b-1$$

So $$\frac{A}{B} =b-2 + \frac{(b-1)^3}{b^b-(b^2-b+1)}$$ and you have that $$\frac{(b-1)^3}{b^b}<\frac{(b-1)^3}{b^b-(b^2-b+1)}<\frac{(b-1)^3}{b^b} + \frac{1}{b^{2b-5}}$$

So, at least for $b$ large enough, you get about $b-3$ zeros followed by the digits of $(b-1)^3$. So the quotient $\frac{987654321}{123456789}\approx 8.00000007290000$, the fact that $729=9^3$ is not some numerical accident.

For instance, we know when $b=8$ that:

$$\frac{(7654321)_8}{(1234567)_8}\approx (6.00000527...)_8$$ since $(527)_8=7^3$.

Answer 6

Consider the product

$$9\cdot123456789=1111111101.$$

This pattern is due to the fact that $9$ is one less than the basis of the numeration so that the products with individual digits (from the right $81,72,63,54\cdots$) have an increasing unit digit, while the tens digit increases. Thus thanks to carries, the digits of the product remains constant (with an anomaly). More specifically, the sum of units and carry from the previous is $10$.

On the other hand

$$9\cdot987654321=8888888889,$$ with a similar effect, and the sum of units and carry from the previous is $8$.

Thus the ratio

$$\frac{987654321}{123456789}$$ is very close to $$\dfrac81.$$

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Addendum:

If you consider the rational $$\dfrac{10}{81}=0.123456790123456790\cdots$$ where the $8$ are overwritten by carries, you observe the more "regular" product

$$9\cdot123456790=1111111110.$$

And considering

$$8\cdot987654320=987654320$$

we have

$$9\cdot987654320=8888888880.$$

Now we have your true coincidence,

$$\frac{987654320}{123456790}=8.$$

If you prefer,

$$\frac{987654321-1}{123456789+1}=8.$$