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?
 A: 2178*4 = 8712 
Consider $E$ cannot be 1. otherwise $D=A$.If $E\ge 5$, then $A=1$, $D=E$, contradiction!
So $E\ge 2$, thus we have 


*

*$E = 2$, then $A$ can only be $1,2,3,4$, since the product should be
4-digit number.
$DE$ has end digit as $A$, then $A$ is even, $A$ can be 2,4, since $A\neq E$, therefore $A=4$. Then $D = 7$($D\neq E$), but we look at the first digit of $\overline{DCBA}$, we should know that $D\ge E*A = 8$, contradiction!

*$E=3$, then $A$ can be $1,2,3$, and we exclude $1,3$ for the same reason. $A=2$, thus $D = 4$, But we expect $D\ge E*A = 6$.

*$E=4$, thus $A = 2$, $DE$ has end digit as $2$, thus $D=8$ or $D=3$, and $D\ge E*A =8$, THUS $D=8$.


Then we solve the other two unknowns by solving an equation:
$8000+400B+40C+32 = 8000+100C+10B+2 $,
which is
$39B+3 = 6C$, where we can see $B\le 1$, since $C\le 9$.
we have $(B,C) = (1,7)$, ONLY.
A: 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:
$2178 \cdot 4 = 8712$. 
