digit problem

Find all natural numbers, for which, if you knock off the first digit, the remaining number is a divisor of the original number.
You can find these numbers here: https://oeis.org/A292683

First of, let us write our number as
$y*10^n + x$ with the extra conditions $y<10$ and $x< 10^n$.
$x$ is obviously the number with the first digit removed, so we have to find all $x$, $y$ and $n$ such that there exists a natural number $z$ such that:
$zx = y*10^n+x$

$(z-1)x = y*10^n$

As the left and right side of the equation have the same prime decomposition, we can find 4 new numbers, $a,b,c$ and $d$ such that:
$\begin{aligned} ab&=x\\cd&=z-1\\ac&=y\\bd&=10^n \end{aligned}$

The last one: $bd=10^n$ can for the same reason be simplified to $b=2^k5^l$ and $d=2^{n-k}5^{n-l}$, simplifing everying to:
$\begin{aligned} x&=a2^k5^l\\z&=c2^{n-k}5^{n-l}+1\\y&=ac \end{aligned}$
We are uninterested in the value of z, so this mean that our original value is:
$ac10^n+a2^k5^l$
with the extra conditions of $ac < 10$ and $n > \log_{10}(a2^k5^l)$
The second condition can be smoothed by replacing $n$ by $m+\lceil{\log_{10}(a2^k5^l)}\rceil$
As all numbers are natural numbers anyway. This reduces the expression to

$ac10^{m+\lceil{\log_{10}(a2^k5^l)}\rceil}+a2^k5^l$
$m,k$ and $l$ can assume any natural number, $a$ and $c$ can assume the following pairs:
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(1,7),(1,8),(1,9),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1),(6,1),(7,1),(8,1) or (9,1).

Each combination generates a unique number, and each number is generated by this formula.