A number theory problem by Jack Rawlin

We know that

$\overline{ABC} - \overline{CBA} = -198$

$\overline{ABC} = 100A + 10B + C$

So the equation can be re-written to be

$99A - 99C = -198$

Which then gives us an equation for $A$

$A = C - 2$

Next we should find out $A$ and $C$ so since we know that

$\frac{\overline{DBAC}}{5}$

is an integer we can conclude that $C$ is a multiple of $5$ below $10$ so it can only be $0$ or $5$ so let's put those values into the equation we have for $A$ to see which value works

$A = (0) - 2 = -2$

That doesn't work since we know that $A > 0$

$A = (5) - 2 = 3$

So $A = 3$ and $C = 5$ .

Onto the second equation given

$\overline{ABC} + B = 327$

$100A + 10B + C + B = 327$

$100(3) + 11B + (5) = 327$

$11B = 327 - 305$

$B = 2$

We now know $A$ , $B$ and $C$ , but what about $D$

$D = ((3) + 2(2) + 3(5)) - 14 = 8$

So $\overline{DBAC} = 8235$

Which then means that

$\frac{\overline{DBAC}}{5} = 1647$