If $ABCDE\times4 = EDCBA$ where $A, B, C, D, E$ are all distinct. Why can't $E$ be $3$?

Question

If $ABCDE\times4 = EDCBA$. Here $A, B, C, D, E$ represent distinct nonzero digits and $ABCDE$ and $EDCBA$ are five digit numbers. Why can't E be three?

Answer 1

Assume $E=3$, then $ABCD3 * 4 = A'B'C'D'2$, so for that to equal $EDCBA$, $A$ would need to be 2. But if $A$ is 2, the first digit of $ABCDE*4$ is greater or equal to 8, which is a contradiction to it being $EDCBA$, with $E=3$ as the first digit.