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

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