"American Billions" puzzle: Find a 10-digit number (no repeats), where the first $n$ digits give a number divisible by $n$, for all $n = 1,\ldots,9$

Question

I have no idea why the problem is apparently called this but here it is.

Using only the digits $0$, $1$, $2$, $\ldots$, $9$, find a ten-digit number such that the number up to the $n$-th digit is divisible by $n$, for all $n\in\{1,\ldots,9\}$.

In other words, the first digit is divisible by $1$ (easy); the number made from just the first two digits is divisible by $2$ (again, easy - just make the second number even); and so on. Since there are ten digits and only nine are needed to satisfy the divisibility rule, there should be one left over at the end.

The closest I have gotten so far is the number $309258641$. $$\begin{align} 3 &= 1 \times 3 \\ 30 &= 2 \times 15 \\ 309 &= 3 \times 103 \\ 3092 &= 4 \times 773 \\ 30925 &= 5 \times 6185 \\ 309258 &= 6 \times 51543 \\ 3092586 &= 7 \times 441798 \\ 30925864 &= 8 \times 3865733 \\ 309258641 &= \color{red}{\text{not divisible by $9$}} \\ 309258647 &= \color{red}{\text{not divisible by $9$}} \end{align}$$

However, this was obtained by trial and error (following some simple rules like the second digit must be even, and the first, second and third digits must add to a multiple of $3$, etc.).

My question is:

Does anyone know of a systematic way of finding the answer, or showing that there isn't one?

Answer 1

By brutish force, I find four solutions: $$3816547209, 3816547290, 7832041659, 8016547239$$

Answer 2

If you want a ten digit solution, $0$ must be the tenth digit and $5$ must be the fifth. The other evens have to go in the even places. Each block of three digits must be divisible by $3$. That means that either each block of three has all three digits the same $\bmod 3$ or has a $0$, a $1$, and a $2$. The fourth and eighth place must be $2$ and $6$ (in either order) because they are preceded by an odd digit and the number to that point needs to be divisible by $4$. The second and sixth places are then $4$ and $8$ (in either order). This reduces the number of possibilities considerably, but there is still a bunch of trial and error.