That depends how exhaustive you are willing to be. The multiplier can only be in the range $2-9$, which is only eight cases. To check $4$ we write $abcd \times 4=dcba$ where concatenation indicates different digits. $a$ has to be even as the last digit of the product, and has to be $2$ or there would be a carry. Then $d$ has to be $8$ so that it can produce $a$. $b$ has to be $0,1,2$ to avoid a carry and is odd because of the carry from the ones place, so it is $1$. Then $c$ has to be $7$ and we are done-$2178 \times 4 = 8712$. $5, 6$ and $8$ are out because $a$ would have to be $1$ to avoid a product over $10,000$ but that is odd and not $5$. $7$ is out because again $a$ would have to be $1$, but then $d$ can't be $3$. So we just have to check $2, 3,$ and $9$-not too much work.