Please help me to find the value of $ABCDE$ (step by step) $ABCD\times E = DCBA$ with $A,B,C,D$, and $E$ distinct decimal digits (and $ABCD$ representing the concatenation of those digits).  How can I find the value each of them?

First of all, note that $ABCD$ and $DCBA$ have the same digit-sum; this means in particular that they both have the same value $\mod 9$ (call it $r$).  But working $\mod 9$, this means that $r\cdot E = r$, meaning that either $E\equiv 1$ (in which case we must have $E=1$ and then trivially $A=D, B=C$ and the solution is invalid) or $r\equiv0$ — in other words, $ABCD$ (and likewise $DCBA$) is a multiple of 9.
Next, we can use the 'alternating digit sum', or in other words work$\mod 11$ : $ABCD\equiv D-C+B-A\equiv q$, say, and $DCBA\equiv A-B+C-D\equiv -q$.  So $E\cdot q \equiv -q\bmod 11$, leaving the possibilities $E\equiv-1$ (impossible, since the first positive number $\equiv -1 \bmod 11$ is $10$) or $q\equiv0$ — in other words, $ABCD$ (and likewise $DCBA$) is also a multiple of $11$, and so in fact it's a multiple of $99$.
With this in hand, we can look at $DCBA$ : since we know it's a multiple of $99$, and in fact a multiple of $99$ by a 2-digit number, we can write $DCBA=99\cdot XY$.  But $99\cdot XY = (100-1)\cdot XY = XY00-XY = XZ\bar{X}W$ where $Z=Y-1$, $W=10-Y$, and $\bar{X}=9-X$ (note that $Y$ can't be $0$ since otherwise that would leave $A=0$, but $A$ is also the first digit of a number), so $D=X, C=Y-1, B=9-X, A=10-Y$.  Likewise, we have $ABCD=99\cdot ST$ (with $ST\cdot E = XY$) yielding $A=S, B=T-1, C=9-S, D=10-T$.  In other words, $E\cdot ST = XY$ with $X=10-T, Y=10-S$.
Now work mod 11 again: $XY\pmod{11}\equiv Y-X\equiv (10-S)-(10-T)\equiv T-S\equiv ST$.  But since $XY=E\cdot ST$, then either $E\equiv 1\bmod 11$ (which we can rule out) or $XY\equiv ST\equiv 0$ - in other words, $XY$ and $ST$ are both multiples of $11$, i.e. $X=Y$ and $S=T$.  Now it's a simple trial-and-error of finding $S$ and $E$ such that $E\cdot S = X = 10-S$; since $E\cdot S$ must be a single digit, there are only a couple of cases to consider and it's easy to find $S=2, E=4, X=8$ (the only alternative, $S=1, E=9, X=9$ gives $D=X=9=E$, so it can be ruled out), leading to $A=2, B=2-1=1, C=9-2 = 7, D=10-2 = 8$ and the final result:
$2178 \cdot 4 = 8712$. 
