Right now, I am studying number theory and I stumbled upon this number:
$$n=\overline{a_k a_{k-1}\dots a_3 a_2 a_1}$$
I found it in this statement: Number $n$ is divisible by 7 $\iff$ the number $m = \overline{a_9a_8a_7} - \overline{a_6a_5a_4} + \overline{a_3a_2a_1} - \dots$ is also divisible by $7$.
Can someone please explain to me what kind of a number is this? Thanks!
According to your comment yes, in this context, this can be used to denote the digits of $n$.
$$n=\overline{a_k\cdots a_2a_1}\iff n=\sum\limits_{i=1}^k a_i\cdot 10^{i-1}$$
Notice that $1000=142\times 7+6$ therefore $\begin{cases}1000\equiv 6\pmod 7\equiv -1\pmod 7\\1000^2\equiv 1\pmod 7\end{cases}$
Let's rewrite our number under the following form:
$\begin{align}n &=\overline{a_9a_8a_7a_6a_5a_4a_3a_2a_1}\\\\ &=\overline{a_9a_8a_7}\times 1000^2+\overline{a_6a_5a_4}\times 1000+\overline{a_3a_2a_1}\\\\ &\equiv\overline{a_9a_8a_7}\times (1)+\overline{a_6a_5a_4}\times (-1)+\overline{a_3a_2a_1}\pmod 7\end{align}$
Which is the desired expression in your comment, $n$ is divisible by $7\iff\overline{a_9a_8a_7}-\overline{a_6a_5a_4}+\overline{a_3a_2a_1}$ is divisible by $7$.
Here are two examples
$n=464\,981\,517$ then since $464-981+517=0$ then $n$ is divisible by $7$.
$n=613\,349\,331$ then since $613-349+331=595=85\times 7$ then $n$ also divisible by $7$.