Divisibility of a series

Question

I use the notation $123 \dots (z)$ to represent a number that looks like a concatenated string of consecutive integers up to $z\in \mathbb{N}$. E.g. $123 \dots (15)$ denotes $12346789101112131415$.

I have observed the following pattern:

If $p|123\dots (p)$ , $q|123\dots (q)$ , gcd(p,q)<=2, then $pq|123\dots (pq)$.

Excluding p or q is 69, is this true.For example $2|12$ and $3|123$, so $6|123456$. Can you prove or disprove the statement? Additionally, what is the rule when $p=q$?

Also, if my aim is to determine whether n|123...(n), what can we do to those n is a prime or 8|n

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P.S. I'm sorry i made so much mistakes. this is a question i investigate on long time ago and it suddenly comes up to me today, I'm sorry I have forgotten some of the details

Answer 1

The numbers $n$ such that $n$ divides $12\dots(n)$ are tabulated at http://oeis.org/A029455. $3$ and $27$ are in the list, but $3\times27=81$ is not, so it's a counterexample. So is $p=9$, $q=27$; also, $p=18$, $q=27$, and $p=27$, $q=36$, and doubtless many others.

Perhaps more interesting: if OEIS is right, then $p=69$ works, but $2\times69$ doesn't, neither $3\times69$, $5\times69$, and so on.

Answer 2

Not exactly the smallest counterexample, but $$ 12345678910111213141516171819202122232425262728293031323334353637383940 $$ is not divisible by $40$ even though $1234567891011121314151617181920$ is divisible by $20$ and $12$ is divisible by $2$.