Why u dun liek mudkipz?

$\large\begin{array}{c}&&T&R&E&E&C&K&O \\ +\ \ &T&O&R&C&H&i&C \\ \hline \ \ S&T&a&R&T&E&R&S \end{array}$

Terminology : I read the cryptogram from left to right, so I will refer to the leftmost column as the first column, and so on...

From the second column, note that without carry over from the third column, we will have $2T \equiv T \pmod{10}$ , possible only if $T=0$ , but we know $T\neq0$ since it is the leading number of both $TORCHiC$ and $TREECKO$ . Therefore, with carry over, $2T+1 \equiv T \pmod{10}$ , or $T \equiv 9 \pmod{10}$ , which means $T=9$ . This also implies that $S=1$ .

We shift our attention to the fourth column. Note that $E+R$ gives $R$ . We can immediately conclude that $E=0$ and there is no carry over from the fourth column. Why? Suppose that there is a carry over. Then, $E+R+1 \equiv R \pmod{10}$ or $E=9$ . This creates a contradiction, since already $T=9$ .

Our partially solved cryptogram now looks like this:

$\large\begin{array}{c}&&9&R&0&0&C&5&O \\ +\ \ &9&O&R&C&H&i&C \\ \hline \ \ 1&9&a&R&9&0&R&1 \end{array}$

We shift our attention to the fifth column, where $0+C$ gives $9$ . Since $9$ is already exhausted, we can only let $C$ be $8$ and conclude that there must be a carry over from the sixth column.

From the last column, clearly, $O=3$ . What remains are the integers $2,4,6,7$ . If we take a look at the third column, we can conclude that $R=7$ in order for a carry over to the second column to be possible. An immediate consequence of this is that $a=0$ .

The rest of the puzzle is now easy to complete. It can be easily seen that $H=2$ and $i=1$ , bearing in mind that there is a carry over from the last column.

A quick computation reveals that $\overline{TREECKO}-\overline{TORCHiC}$

$=9700853-9378218$

$=\boxed{322635}$