Find ABBC

Brute force/coding would work fine here, but there is a lot we can deduce first.

Firstly, $6x<10000$ so $x<1667$ , and $A$ must be $1$ . Also $B<7$ .

The first digit of $y$ must be one of $6,7,8,9$ . But this is also its last digit; $y$ is a multiple of $6$ so it must be even; so $D$ is either $6$ or $8$ .

If $D$ were $6$ , we would have $y<7000$ and so $x<1167$ ; so $B$ would have to be $0$ or $1$ . But neither of these is allowed. So $D=8$ .

$6C \equiv 8 \pmod{10}$ ; so $C=3$ or $C=8$ . We can't have both $C$ and $D$ equal to $8$ ; so $C=3$ .

This leaves as the only possibilities for $x$ the following: $1223$ , $1443$ , $1553$ , $1663$ . It's now easy to check by hand that the only one of these that gives the correct form for $y$ is $x=\boxed{1443}$ .