A father wants to divide his property among his children: first, he gives 1000 yuan and one-tenth of the remaining property to the eldest child, then 2000 yuan and one-tenth of the remaining property to the second child, then 3000 yuan and one-tenth of the remaining property to the third child, and so on. It turns out that each child receives the same amount of property. The father has $\qquad$ children.
(Method 1)
Let the father's property be $x$ yuan;
Since the eldest and the second son receive the same amount of property, we have
$$
1000+(x-1000) \times \frac{1}{10}=2000+\left[(x-1000) \times\left(1-\frac{1}{10}\right)-2000\right] \times \frac{1}{10} ;
$$

Solving this, we get $x=81000$;
The eldest son receives $1000+(81000-1000) \times \frac{1}{10}=9000$ yuan;
That is, each child receives 9000 yuan;
This father has a total of $81000 \div 9000=9$ children.
(Method 2)
Since each child receives the same amount of property;
The eldest and the second son receive the same amount of property;
The eldest son receives 1000 yuan plus $\frac{1}{10}$ of the remaining property (let's denote this remaining as $A$); the second son receives 2000 yuan plus $\frac{1}{10}$ of the remaining property (let's denote this remaining as $B$); thus, $\frac{1}{10}$ of $A$ is 1000 yuan more than $\frac{1}{10}$ of $B$;
$A$ is $1000 \div \frac{1}{10}=10000$ yuan more than $B$;
From $A$ to $B$, it decreases by $\frac{1}{10}$ of $A$ and 2000 yuan;
Therefore, $\frac{1}{10}$ of $A$ is $10000-2000=8000$ yuan;
$A$ is $8000 \div \frac{1}{10}=80000$ yuan;
This father's property is $1000+80000=81000$ yuan;
The eldest son receives $1000+(81000-1000) \times \frac{1}{10}=9000$ yuan;
That is, each child receives 9000 yuan;
This father has a total of $81000 \div 9000=9$ children.
(Method 3)
Let this father have $n$ children;
Then the second-to-last child receives $1000(n-1)$ yuan plus $\frac{1}{10}$ of the remainder;
The last child receives $1000n$ yuan;
These $1000n$ yuan are $\frac{9}{10}$ of what remains after the second-to-last child receives $1000(n-1)$; each child receives the same amount of property;
From the equality of the property received by the second-to-last and the last child, we have
$$
1000(n-1)+1000n \times \frac{1}{9}=1000n \text {; }
$$

Solving this, we get $n=9$;
This father has a total of 9 children.