Find CAB

$\begin{array} {rrrr} & \blue A & B & \red C \\ - & \blue C & B & \red A \\ \hline & \blue C & A & \red B \end{array}$

Since $A$ , $B$ , and $C$ are positive integers, from the hundredth column (blue) we know that $A > C$ . In fact $A \ge 2C$ . This means that from the unit column (red), $\red 10+C - A = B$ with $\red 1$ borrowed from $B$ of $\overline{ABC}$ . And the tenth column becomes $10+B - 1 - B = 9 \implies A = \red 9$ as follows:

$\begin{array} {rrrr} & \red 9 & B & C \\ - & C & B & \red 9 \\ \hline & C & \red 9 & B \end{array}$

From unit row $10+C - A = B \implies \blue{B = C+1}$ . Then we have:

$\begin{aligned} 900 +10B + C - 100C - 10B - 9 & = 100 C + 90 + B \\ 199C + B & = 801 & \small \blue{\text{Note that }B=C+1} \\ 200C + 1 & = 801 \\ \implies C & = 4 \\ B & = C+1 = 5 \end{aligned}$

Therefore, $\overline{CAB} = \boxed{495}$ .

We have $(100A+10B+C)-(100C+10B+A)=(100C+10A+B)$ . Tidying this up, we get $89A=199C+B$ .

Divide through by $89$ to get $A=2C+\frac{21C+B}{89}$ . So $21C+B$ is a multiple of $89$ . Since both $B$ and $C$ are single-digit numbers, the largest the left-hand side can possibly be is $21 \times 9 + 9 = 198 < 3 \times 89$ .

So either $21C+B=89$ or $21C+B=178$ (assuming $C>0$ ; if we don't make this assumption we also get the trivial solution $CAB=000$ ). It's easy to see the only solution in single-digit numbers is $C=4$ , $B=5$ ; substituting back in to the original equation, we get $A=9$ ; so $CAB=\boxed{495}$ .