How many possible combinations of 9 digits are there as long as they follow a certain set of rules? The digits 1-9 are labelled a1, a2, a3, ..., a9 such that:
a1 is a multiple of 1.
a1a2 is a multiple of 2 (here and throughout the digits are concatenated rather than multiplied - so if a1 is 4 and a2 is 3 then
a1a2 = 43 and not 12).
a1a2a3 is a multiple of 3.
a1a2a3a4 is a multiple of 4.
And so on until a1a2a3a4a5a6a7a8a9.


*

*Can you find a suitable arrangement of the digits?

*How many possibilities are there?


I began forming a few rules, so I had:
a2, a4, a6, a8 must all be even.
a5 must be 5.
a1 + a2 + a3 = multiple of 3.
a4 + a6 = 10.
Using these rules, I managed to answer the first question and got the arrangement 381654729.
For Part 2, I was thinking of listing them all out, but I was hoping there is a shorter and faster method to approach this.
 A: To cut down on how many cases you need to examine note that since $8\mid a_1\dots a_8$, $8\mid a_6a_7a_8$. This, together with the distinct-digits condition, alternating even and odd digits ($a_{\text{odd}}$ is odd, $a_{\text{even}}$ is even) and $a_7\ne5$, restricts $a_6a_7a_8$ to $12$ possibilities:
$$216,296,416,432,472,496,632,672,816,832,872,896$$
But now note that $a_6$ cannot be $2$ or $6$, as that would make $a_4$ $4$ or $8$. Since $4\mid a_1a_2a_3a_4$, $4\mid a_3a_4$ and $a_4$ being $4$ or $8$ would force $a_3$ to be even, which it cannot be. The $12$ possibilities are reduced to $8$:
$$416,432,472,496,816,832,872,896$$
Then of course $a_6+a_8\ne10$, otherwise $a_4=a_8$. We go down to four possibilities:
$$432,472,816,896$$
Now suppose $a_6=8$. We get the partial $?4?258?6?$; since $1,7,4$ are $1\bmod3$ and $3,9$ are $0\bmod3$, $a_1,a_7$ must be $1,7$ in some order, $a_7=9$ and $a_9=3$. But neither $1472589$ nor $7412589$ are divisible by $7$, so any solution satisfies the following partial:
$$?8?654?2?$$
Casework should now be small enough to do by hand and show that $381654729$ is the only solution.
