The Peculiar License Plate

Let the original license plate number be $abcde$ and the reversed one $ABCDE$ . We know that the numbers readable in reverse are $0,1,6,8,9$ . We are given that $ABCDE-edcba=78633$ or:

$\begin{matrix} \\ - \\ \end{matrix}\begin{matrix} A \\ e \\ 7 \end{matrix}\begin{matrix} B \\ d \\ 8 \end{matrix}\begin{matrix} C \\ c \\ 6 \end{matrix}\begin{matrix} D \\ b \\ 3 \end{matrix}\begin{matrix} E \\ a \\ 3 \end{matrix}$

Let's look first pair of digits $A$ and $e$ . For the difference to be $7$ , there are only two cases, $(A, e)=(9,1)$ or $(8,1)$ . If $(A,e)=(9,1)$ , then the last pair of digits are $(1,6)$ and $....1-....6 \ne ....3$ , therefore, $(A,e)\ne(9,1)$ and $(A,e)=(8,1)$ . And we note that $....1-....8=....3$ .

Similarly, for .(B...-.d...=.8...), $(B,d)\ne(8,0)$ because $(...D.-...b. =...1.$ instead of $...3.$ .. Therefore, $(B,d)=(9,0)$ and $(D, b)=(0,6)$ , and we note that $...0.-...6.=...3.$ .

Now, the remaining $(C,c)$ . $(C,c)\ne(0,0)\ne(1,1)\ne(8,8)$ , else $..C..-..c..=..9..$ instead of $..6..$ . $(C,c)=(6,9)$ fits perfectly.

Therefore, the answer is $89601-10968=78633$ , and the original plate number is $\boxed{10968}$ .

Readable numbers in reverse positions: 0, 1, 6, 8, and 9

Increase of 70k something --> must be 8 and 1 (difference of 7)

So I put it: 8xxx1 - 1xxx8 = 78633

Then I looked on the 10th value. The difference is 3, but remember we have taken 10 previously to make 1(0) minus 8 becomes 3. So the difference should be 4. It must be 0 and 6.

So I put it: 86x01 - 10x68 = 78633

Now there are only one pair of numbers left --> with difference of 6. Since we have taken 1 from 10th value to make zero minus six becomes 3 as well; the only possible combination is 6 and 9.

Therefore: 86901 - 10968 = 78633

We can do a check and find the equation is correct.

10968

+78633

89601

Which is seen when it is up side down....