<s>[INST] 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!
 [/INST] 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
$$
\begin{align}
37 \cdot 111\cdots 1 \quad (n \text{ ones}) &=
37u_n\\&= 36u_n + u_n\\
&= 4\cdot9\cdot u_n+10u_{n-2}+11\\
&=4(10^n-1)+10u_{n-2}+4+7\\
&=4\cdot10^n+10u_{n-2}+7\\
&=4111\cdots107 \quad (n-2 \text{ ones})
\end{align}
$$
</s>