Find A,B,C,D consecutive numbers based on written addition formula

Question

$A, B, C, D$ are consecutive digits: $B$ is greater by $1$ than $A$, $C$ is greater by $1$ than $B$, $D$ is greater by one than $D$. Four $X$s are digits $A,B,C,D$ in unknown order. Find $A,B,C,D$ \begin{matrix} & A & B & C & D\\ & D & C & B & A\\ +& X & X & X & X\\ \hline 1& 2 & 3 & 0 & 0\\ \end{matrix}

The only thing I can think about is assign $A = y, B = y+1, C = y+2, D=y+3$ and then insert it above.

Not homework.. I try to prepare myself.

EDIT: after assignment I get: \begin{matrix} & 2y+3 & 2y+3 & 2y+3 & 2y+3 \\ + & X & X & X & X \\ \hline 1 & 2 & 3 & 0 & 0\\ \end{matrix}

So:

\begin{matrix} & 2A+3 & 2A+3 & 2A+3 & 2A+3\\ + & X & X & X & X\\ \hline 1 & 2 & 3 & 0 & 0\\ \end{matrix}

Answer 1

$ABCD+DCBA = 1111\cdot(2A+3)$

If $A=1$, $12300-5\cdot1111=6745$.

If $A=2$, $12300-7\cdot1111=4523$.

If $A=3$, $12300-9\cdot1111=2301$.

The next three possibilities will be too small, as we need a four digit number. So $A=2$.