TITLE: Why does adding a suitable multiple of $9$ always lead to the reverse of the number?
QUESTION [50 upvotes]: For example: 

$12$ reversed is $21$ and $12$ + $9$ = $21$. 
$17$ with the two values swapped is $71$, and $17$ + $9$ + $9$ + $9$ + $9$ + $9$ + $9$ = $71$. 
Take the number $123$ and add $9$ a total of $22$ times you get $321$, which is the reverse. 

It seems to work for every number. Why is this the case? Is it just a addition thing?

REPLY [3 votes]: Lemma: $\overline{ab}-\overline{ba}\equiv 0 \pmod 9$.
  Proof: $(10a+b)-(10b+a)=9(a-b)\equiv 0 \pmod 9$.

Based on the Lemma, $\overline{abc}-\overline{cba}\equiv 0\pmod 9$, because:
$$\overline{abc}-\overline{acb}\equiv \overline{acb}-\overline{cab}\equiv \overline{cab}-\overline{cba}\equiv 0 \pmod 9 \iff \\
(\overline{abc}-\overline{acb})+(\overline{acb}-\overline{cab})+(\overline{cab}-\overline{cba})\equiv 0 \pmod 9 \iff \\
\overline{abc}-\overline{cba}\equiv 0\pmod 9.$$
Note: It is $T(2)=3:\overline{abc}\Rightarrow \color{red}{\overline{acb}\Rightarrow \overline{cab}\Rightarrow \overline{cba}}$.
In general, $\overline{a_1a_2...a_n}-\overline{a_na_{n-1}...a_1}\equiv 0 \pmod 9$ and: 
$$\overline{a_1a_2...a_n}\overbrace{\Rightarrow}^{T(n-1)} \overline{a_na_{n-1}...a_1}.$$