$4\times ABCDE = EDCBA$: Four times a five digit integer is that integer backwards. A student gave me this puzzle the other day. Where $A,B,C,D,E$ are distinct digits, and where $A,E\ne0$, what 5 digit integer satisfies the condition below?
$$4\times ABCDE=EDCBA$$
What I'm interested in isn't the answer per se. I'm looking for the most efficient solution - the least number of logical moves necessary to solve the problem.

Clearly $A$ is even and so $A=2$ to avoid a carry in multiplication, so $E≥4$ is $8$ or $9$ but $4×9=36$ so $E=8$ (credit to the comment on the question).  $B<3$ for $E=8$ . $B$ is odd since $B=4D+3$ where $3$ is the carry. So, $B=1$  and $D=2$ or $7$  If $D=2$, carry under $B$ should have been $8$ which is impossible with multiplication by $4$. So, $D=7$  Now, it's easy to derive $4C+3=3*10+C$ and $C=9$ .  The number is $21978$
