Does every positive integer appear in the digits of $2\cdot 0.1234567891011… $? Let $C = 0.1234567891011121314…$ the Champernowne constant. My question is :

Does the real number $2 \cdot C \simeq 0.24691357820222426283032343638404244464850525456586062646668707274...$ contain every positive integer in its digits?

For instance, $2022$ appears here : $0.246913578\underbrace{2022}...$
Obviously this is true for $C$, and also for $0.246810121416...$. These number are known to be normal.
My question is to determine whether $2C$ is at least a disjunctive number (in base $10$).
More generally, I would like to know if a non-zero multiple $n \cdot x \; (n \in \Bbb Z)$ of a disjunctive number is also disjunctive (true if $n=10^k,k\in \Bbb N$).
I looked at some theorems about disjunctive/normal numbers (for instance, if $f$ is a non-constant polynomial with real coefficients which is always positive, then the "concatenation" of the integer parts $x=0.[f(1)][f(2)][f(3)]...$ is normal), but I wasn't able to conclude.
Any comment would be helpful.

In general, if you want to show that an $m$ occurs in $nC$, then write $m/n$ in decimal to a large amount of digits, then round up at the end. Then when that sequence of digits occurs in $C$ surrounded by "enough" zeros, then $m$ occurs in the same location of $nC$. 
(When writing $m/n$, you'll have to keep the zeros that start the quotient in the case $n>m$.)
For example, given $m=73$, $n=6$ then $m/n=12.16666\dots$ so we pick the digits $12167$. Then where $10121670$ occurs in $C$, $6073002$ occurs in $6C$.
This is particularly easy for $n=2^k$ or $n=5^k$ because the decimals for $m/n$ terminate in these cases, so there is no question of "how many digits." I think, more generally, you are safe if you take more digits of $m/n$ after the decimal than there are digits of $n$, but I'm not 100% sure of that.
You have to start with $1$ then add as many digits of $0$s as digits of $n$, then add your sequence of digits from $m/n$, then again as many $0$s to the end as digits of $n$.
So if $m=84,n=13$ then $m/n \approx 6.461538462$ so we'll take the digits $10064700$ and find them in $C$. Where these digits occur, we'll get $308411xx$ in the same location of $13C$.
The only property of $C$ we are using is that every finite sequence of digits occurs in it.
