Numbers equal to the arithmetic mean of their permutations 
An $n$-digit number $\alpha$ is said to be special if $\alpha$ is equal to the arithmetic mean of all the permutations one obtains by rearranging the digits of $\alpha$ in all possible ways, and the digits of $\alpha$
  are not all equal. Prove that any special number $\alpha$ must have exactly $3k$ digits, where $k$ is a positive integer.

Note that this is https://oeis.org/A161020/internal.
I thought a proof by contradiction might work.
Assume, for sake of contradiction, that a special number $\alpha = \overline{a_{3k+1} \ldots a_{1}}$ existed with the required property with exactly $3k+1$ digits where the digits are distinct. Then the arithmetic mean of the permutations of the number is $$\dfrac{\underbrace{(3k)!(3k)!\ldots(3k)!}_{3k+1}(a_{3k+1}+\cdots+a_1)}{(3k+1)!}=10^{3k}a_{3k+1}+10^{3k-1}a_{3k}+\cdots+a_1$$
I wasn't sure what to do next.
 A: The $2$'s in this are wrong:
$$\dfrac{\underbrace{22\ldots2}_{3k+1}(a_{3k+1}+\cdots+a_1)}{(3k+1)!} = 10^{3k}a_{3k+1}+\cdots+a_1$$
and it's not an identity.
It should be:
For what $a_{3k+1}a_{3k}\dots a_2a_1$ does this equation hold true:
$$\dfrac{\underbrace{(3k)!(3k)!\ldots(3k)!}_{3k+1}(a_{3k+1}+\cdots+a_1)}{(3k+1)!}=10^{3k}a_{3k+1}+10^{3k-1}a_{3k}+\cdots+a_1$$
with the number that is under-braced being treated appropriately.
The $(3k)!$ can be seen by example, from considering say the case $k=1$ and $1234$. Arrange each permutation above one another, then each column sums to $10^c\times6\times(1+2+3+4)$, with $c$ representing the column number starting on the right with $c=0$ and finishing on the left with $c=3$.
The final number for the case $k=1$ and general number $abcd$ is $6666(a+b+c+d)$.
With larger $k$ we consider $\sum\limits_{i=0}^{3k}10^i(3k)!=(3k)!\sum\limits_{i=0}^{3k}10^i$.
And this means that in the completely general case:
$$\dfrac{\underbrace{(k!)(k!)\ldots(k!)}_{k+1}}{(k+1)!}$$
becomes:
$$\dfrac{\sum\limits_{i=0}^k 10^i}{k+1}$$
So your next step is to determine when:
$$\dfrac{\sum\limits_{i=0}^{3k} 10^i}{3k+1}(a_{3k+1}+a_{3k}+\dots+a_1)=10^{3k}a_{3k+1}+10^{3k-1}a_{3k}+\cdots+a_1$$
In the general case, for $a_{k+1}a_k\dots a_1$, the equation is:
$$\dfrac{\sum\limits_{i=0}^k 10^i}{k+1}(a_{k+1}+a_k+\dots+a_1)=10^k a_{k+1}+10^{k-1}a_k+\cdots+a_1$$
