Providing an explanation for trend found in numbers and divisibility. Let $A$ and $B$ be integers with the same number of digits (like 35 and 92). Why is it that $$\underbrace{AA\ldots A}_{W\ A's} \cdot \underbrace{BB\ldots B}_{Z\ B's} = \underbrace{AA\ldots A}_{Z\ A's}\cdot \underbrace{BB\ldots B}_{W\ B's}?$$ (Here the $A$'s and $B$'s are being concatenated, not multiplied.) For example A=3 and B=8, W=2 and Z=4, why would 33*8888=3333*88? Why does this equation work?
I was drinking coffee when I randomly came up with this so I was not quite successful in finding an answer. I tried explaining this trend using the logarithmic graph below. I was trying to see whether changing the difference between W and Z would have an effect on the trend and it didn't. What I noticed did have a trend was the actual smaller number between W and Z. The larger the smallest of the two was, the more the A_W/A_Z function began to behave like a regular logarithm. Can someone try to explain 1- why  the expression even works in the first place and 2- why the division of AW/AN approaches a 10^n where n= the difference between W and N 
 A: (This is too long to post as a comment on the answer by @fractal1729 .)
It might be worth mentioning that the numbers $X$ and $Y$ are given by geometric series: $$X=10^{N(W-1)}+10^{10(W-2)}+...+10^{1N}+10^{0N}=\frac{10^{NW}-1}{10^N-1}\\ Y=10^{N(Z-1)}+10^{10(Z-2)}+...+10^{1N}+10^{0N}=\frac{10^{NZ}-1}{10^N-1}.$$
Let $A^W$ denote the numeral formed by concatenating $W$ copies of $A$, and let $N$ be the length of $A$. By inspection we have
$$\begin{align}A^W &=A^{W-1}10^N + A
\end{align}$$
and by applying this recursively, we find
$$\begin{align}A^W &=A^{W-1}10^N + A\\
&=(A^{W-2}10^N+A)10^N+A\\
&=A^{W-2}10^{2N}+A\,10^N+A\\
&...\\
&=A\cdot(10^{(W-1)N}+10^{(W-2)N}+...+10^{1N}+10^{0N})\\
&=A\cdot\frac{10^{NW}-1}{10^N-1}
\end{align}$$
and similarly for $B^Z$. 
A: The reason is that $$33\cdot8888 = (3\cdot11)(8\cdot1111) = (8\cdot11)(3\cdot1111) = 88\cdot3333.$$  In general, if $A$ and $B$ have $N$ digits, we have $$\underbrace{AA\ldots A}_{W\ A's} = A\cdot \underbrace{1\underbrace{00\ldots0}_{N-1\ 0's}1\underbrace{00\ldots0}_{N-1\ 0's}100\ldots\ldots01\underbrace{00\ldots0}_{N-1\ 0's}1}_{W\ 1's} = A\cdot X.$$  Let $X$ be the number on the right that is multiplied by $A$.  Similarly we define $Y$ as the number on the right of the below equation that is multiplied by $B$. $$\underbrace{BB\ldots B}_{Z\ B's} = B\cdot \underbrace{1\underbrace{00\ldots0}_{N-1\ 0's}1\underbrace{00\ldots0}_{N-1\ 0's}100\ldots\ldots01\underbrace{00\ldots0}_{N-1\ 0's}1}_{Z\ 1's}= B\cdot Y.$$
Putting everything together, we get $$\underbrace{AA\ldots A}_{W\ A's} \cdot \underbrace{BB\ldots B}_{Z\ B's} = (A\cdot X)(B\cdot Y) = (A\cdot Y)(B\cdot X) = \underbrace{AA\ldots A}_{Z\ A's}\cdot \underbrace{BB\ldots B}_{W\ B's}.$$
