I assumed $n=3$ is actually a typo and $n\geq3$ is what you meant to say.
Let $\overline{a_2a_3...a_n}=c$ and $a_1=a$.
$\Rightarrow\overline{a_1a_2a_3...a_n}=10^{n-1}a_1+c=10^{n-1}a+c$.
$$c\mid 10^{n-1}a+c\Rightarrow c\mid 10^{n-1}a=2^{n-1}5^{n-1}a$$
But we also have $10\nmid c$, so there are three options. (If $10\mid c$, then $a_n$=0.)
If $2\nmid c, 5\nmid c$, then $c\mid a$ which means $c$ is a "one-digit" number. $n=2$.
If $2\mid c$, but $5\nmid c$, then $c\mid 2^{n-1}a$.
$$\Rightarrow 10^{n-2}<c\leq 2^{n-1}a\leq 9.2^{n-1} \Rightarrow 5^{n-2}<18\Rightarrow n\leq 3$$
If $5\mid c$, but $2\nmid c$, then in a similar manner, we'll get $$10^{n-2}<c\leq 9.5^{n-1}\Rightarrow 2^{n-2}<45\Rightarrow n\leq 7$$
So maximum possible value of $n$ is $7$. But when $n=7$, we get $c\mid 5^6a$
When $a=9$, $c\mid 140625$, $c>10^5$, $\Rightarrow c=140625$. but $a_i\neq 0$.
Similarly, for $a=8$ and $a=7$, we get $c=125000$ and $c=109,375$, both contain $0$, and for $a<7$, we cannot get $c>10^5$.
So $n=7$ is not the answer we want, $n=6$ is.
$$928125=33.28125.$$