Modular guessing?

Since we have $\overline{ABCD} \equiv 11 \pmod{15}$ , we know that 15 divides $\overline{ABCD} - 11$ evenly.

From $A \equiv 1 \pmod{7}$ we know that $A = 1 / 8$ .

From $\overline{AB} \equiv 0 \pmod{7}$ we know that $if~A = 1,~then~B = 4;~if ~A = 8,~then~B =4$ . We conclude that $B = 4$

From $C \equiv D \equiv 1 \pmod{5}$ we know that $C = 1 / 6$ , $D = 1 / 6$

From $C + A \equiv 0 \pmod{7}$ we can work out $C$ . All possible values of $C + A$ are: $1 + 1 = 2;~1 + 8 = 9;~6 + 1 = 7;~ 6 + 8=14$ . We conclude that $C = 6$ .

Now we know that our number looks like this: $\overline{A46D}$ . There are four numbers that can be made up with currently possible values of $A$ and $D$ , however, only one of them is evenly divisible by $15$ when it is subtracted by 15.

Possible numbers are: $1461; 1466; 8461; 8466$ .

Answer: $\boxed{1466}$