Identifying the digits of $37 \cdot aaaa\ldots a$. With a calculator, I have noticed that the integer $37$ multiplied with some particular numbers yields numbers with some structures. 
For instance, let $aaaa\ldots a$ be a natural number of $n$ identical digits. Then, $ 37 \cdot aaaa\ldots a$ is a number with $n+1$ or $n+2$ digits of the form

$$\underbrace{4\cdot a}_{1\text{ or }2}~ \underbrace{aaa\ldots a}_{n-2} ~\underbrace{7\cdot a}_{2^{*}}.$$

$*$ if $a=1$, then $7 \cdot a$ is the sequence of digits $07$. 
I am wondering whether the result above can be proven using some number-theory tool.
Thanks in advance!
 A: Nice observation!
This follows because $aaa\cdots a = a \cdot 111 \cdots 1$ and
$37 \cdot 11 = 407$
$37 \cdot 111 = 4107$
$37 \cdot 1111 = 41107$
$\cdots$
Indeed, let $u_n = 111 \cdots 1$ ($n$ ones). Then $u_n = \dfrac{10^n-1}{9}$ and
$$
\begin{align}
37 \cdot 111\cdots 1 \quad (n \text{ ones}) &=
37u_n\\&= 36u_n + u_n\\
&= 4\cdot9\cdot u_n+10u_{n-2}+11\\
&=4(10^n-1)+10u_{n-2}+4+7\\
&=4\cdot10^n+10u_{n-2}+7\\
&=4111\cdots107 \quad (n-2 \text{ ones})
\end{align}
$$
A: This is not only true for $37$ but all $2$-digit numbers of form $[ab]$, where $a + b =10$.
Since $[ab]=10a+b$
$
\begin{align}
&\ \ \ \ \ [ab]\times11\\
&=(10a+b)\times11\\
&=100a+10(a+b)+b\\
&=100(a+1)+b\\
&=[(a+1)0b]\\
\end{align}
$
In a similar way,
$
\begin{align}
&\ \ \ \ \ [ab]\times11...n\ times...11\\
&=10^{n}a+10^{n-1}(a+b)+10^{n-2}(a+b)...+10(a+b)+b\\
&=10^{n}(a+1)+10^{n-1}+...+100+b\\
&=[(a+1)11...n-2\ times...110b]\\
\end{align}
$
Now for $[ab]\times [kk..n\ times..kk]$
$
\begin{align}
&\ \ \ \ \ [ab]\times11...n\ times...11\times k\\
&=[(a+1)11...n-2\ times...110b]\times k\\
&=[(k\times(a+1))kk..n-2\ times..kk00]+(k\times b)\\
\end{align}
$
This with some modification can be applied to all numbers of form $[abc...]$, where $a+b+c+...=10$.
A: Observe the patterns in the written calculation of these products
$$a=1\to\begin{matrix}3&7\\&3&7\\&&3&7\\&&&3&7\end{matrix}$$
$$a=2\to\begin{matrix}7&4\\&7&4\\&&7&4\\&&&7&4\end{matrix}$$
$$a=3\to\begin{matrix}1&1&1\\&1&1&1\\&&1&1&1\\&&&1&1&1\end{matrix}$$
$$a=4\to\begin{matrix}1&4&8\\&1&4&8\\&&1&4&8\\&&&1&4&8\end{matrix}$$
$$a=5\to\begin{matrix}1&8&5\\&1&8&5\\&&1&8&5\\&&&1&8&5\end{matrix}$$
$$\cdots$$
$$a=9\to\begin{matrix}3&3&3\\&3&3&3\\&&3&3&3\\&&&3&3&3\end{matrix}$$
A digit of the product, in a complete column, is the sum of the digits of $37\cdot a$ plus a possible carry from the previous column, i.e. the sum of the digits of the sum of the digits of $37\cdot a$.
The initial and final digit pairs correspond to partial sums.
We have
$$
37\cdot1=37\to10\to1,\\
37\cdot2=74\to11\to2,\\
37\cdot3=111\to3\to3,\\
37\cdot4=148\to13\to4,\\
37\cdot5=185\to14\to5,\\
\cdots\\
37\cdot9=333\to9\to9.$$
The phenomenon is explained by the fact that $37=4\cdot9+1\to10\to1$, and this occurs for all factors of the form $k\cdot9+1$. For example,
$$73\cdot66666666666666=4866666666666618.$$
