Can any digit be $ 0 $?

My solution: Let the number be $ a.10^n + b $, where $b$ has $n$ digits, and so does $a$. Then $ ab | a.10^n + b \Longleftrightarrow a | b , b | a.10^n $. Now, $a$ and $b$ have the same number of digits so $ b = ma$ where $m$ does not exceed $9$. Hence $ ma | a.10^n \Longleftrightarrow m | 10^n$. So $m$ is either $2$ or $5$. Thus such numbers are of the form $ a(10^n + 2)$, where $a$ is an $n$-digit number less than $5.10^{n-1}$, or of the form $a(10^n + 5)$, where $a$ is an $n$-digit number less than $2.10^{n-1}$. Edit : $m=1$ is also possible. Then such numbers are of the form $ a(10^n + 1)$ where $a$ is an $n$-digit number. Yes, further refining: If the number is of the form $ a(10^n + 2)$ then $ 2a^2 | a(10^n + 2) \Rightarrow a | 5^n + 1 \Rightarrow a = 5^n + 1$ or $ a = \frac{5^n +1}{2} $. Similar refining with others.

ComplexPhi wrote: Find all the positive integers $N$ with an even number of digits with the property that if we multiply the two numbers formed by cutting the number in the middle we get a number that is a divisor of $N$ ( for example $12$ works because $1 \cdot 2$ divides $12$) Let $N=a10^n+b$ with $a$ having exacrly $n$ digits an $b$ having at most $n$ digits. We are looking for $a10^n+b=kab$ and so $b=ma$ and $10^n+m=kma$ So $m|10^n$ and $m<10$ (since $a$ and $ma$ both have $n$ digits) and so $m\in\{1,2,4,5,8\}$ 1) $m=1$ Equation is $10^n+1=ka$ and so $a=b=\frac{10^n+1}k$ which implies $k=7$ (remember $a,b$ have $n$ digits) and $n\equiv 3\pmod 6$ Hence the solutions $\boxed{\text{S1 : }N=\frac{(1000^{2n-1}+1)^2}7\text{ }\forall n\in\mathbb Z_{>0}}$ 2) $m=2$ Equation is $10^n+2=2ka$ and so $b=2a$ and $a=\frac{10^n+2}{2k}$ and so : $k=2$ and $n=1$ hence the solution $\boxed{\text{S2 : }N=12}$ $k=3$ and the solutions $\boxed{\text{S3 : }N=\frac{(10^n+2)^2}6\text{ }\forall n\in\mathbb Z_{>0}}$ 3) $m=4$ Equation is $10^n+4=4ka$ and so $b=4a$ and $a=\frac{10^n+4}{4k}$ and so : $k=2$ and $n=2$ hence the solution $\boxed{\text{S4 : }N=1352}$ 4) $m=5$ Equation is $10^n+5=5ka$ and so $b=5a$ and $a=\frac{10^n+5}{5k}$ and so : $k=3$ and $n=1$ hence the solution $\boxed{\text{S5 : }N=15}$ 5) $m=8$ Equation is $10^n+8=8ka$ and so $b=8a$ and $a=\frac{10^n+8}{8k}$ and so no solution

Maybe further refining is possible in my sol?