Recall from ŧhis question that we call a positive integer slimdownable or slim for short if it is part of a sequence of integers where each is followed by itself divided by its length, i.e. its number of digits. In particular, each must be divisible by its length and the sequence will be falling until it hits a single-digit number.
Examples:
108: slim because $108\overset{/3}{\rightarrow}36\overset{/2}{\rightarrow}18\overset{/2}{\rightarrow}9\overset{/1}{\circlearrowleft}$
78: not slim because $78\overset{/2}{\rightarrow}39\overset{/2}{\rightarrow}\Vert$
Prove or disprove that for any positive integer $n$ there exists a slim number with $n$ digits.
Note: you may use a computer either to produce a counter example or, for example, to complement an asymptotic result, or whatever else you see fit. If you choose to do so, in order to validate your code please answer the following test questions:
Are there any solutions with $11111$ digits? If yes: How many? What are the first 10 digits of their median?
For each $n$ between $1$ and $11111$ calculate the number od solutions with $n$ digits. What is the largest count?
Here is a test case which you can use as a quick sanity check for your code:
At 3590 digits there are four solutions. They all slim down to $6$. First ten digits of these numbers are '3159252337...', '3735860235...', '4606981484...', '6706597705...'.