The Curious Licence Plate

The upside-down number must be at least 10000 bigger than 78633, so its first digit must be 8 or 9 while the first digit of the original number must be 1 or 2. If the first digit or the upside-down number was 9, then the last digit of the original number would have to be 6, so the last digit of the upside-down number would have to be 3+6=9, so the first digit of the original number would have to be 6, which is too large. So the first digit of the upside-down number, and hence the last digit of the original number, are 8, which means the last digit of the upside-down number, and hence the first digit of the original number, are 1.

Now the original number must be less than 90000-78633=11367, so its second digit must be 0 or 1. But we know it has five distinct digits, so the second digit of the original number, and hence the fourth digit of the upside-down number, are 0. By considering the fourth and fifth columns of summation, we now see that the fourth digit of the original number is 6 and hence the second digit of the upside-down number is 9.

Now we have 10?68+78633=89?01, where the two ?'s are each other upside down and therefore must be both 2, both 5, or 9 and 6 respectively (bearing in mind that each number has 5 distinct digits). We can check these 3 cases by hand and verify that 9 and 6 is the only one that works.

The rotated plate has the same quantity of digits with the original, so at least the original has 5-digit number on it.

Listing off the rotational digits, we can divide them into 2 categories, singles or doubles. The singles are 0, 1, 2, 5 and 8 while the doubles are the 6–9 pair.

By the difference of the last digit, the first and last digits are of opposite parity. It could be 1–8 or 8–5 or 5–2 or 9–6, but we already know that the original last digit is bigger than the original first digit which equals the rotated last digit (this from "The value INCREASED by 78633"), so it can only be 1–8. Also by this discovery, the number of digits can only be 5.

So we have now
78633 = 8xxx1 – 1xxx8
Moving on to the second last digits difference, it's a 3 after giving a ten to the ones place (1 borrowed 10 to subtract a bigger 8, remember?). Therefore, we're looking to find a pair with a difference of 4 originally (before the lending), thus the second last digits (comparison between the rotated above and the original below) are of the same parity. This could be 0–6 or 2–8 or 5–1 or 6–2 or 9–5, but also at the same time this corresponds to 9–0 or 8–2 or 1–5 or 2–9 or 5–6 respectively to have a difference of at least 8 without borrowing (since 8 – 1 = exactly 7, nothing sent over). Only possible with 9–0 for the second condition, so our second last digits are 0–6. You can cut on the possible listings a lot if the digits analysis was done simultaneously for the second and fourth digits, but I just decided to be inclusive this time. { You know that a flipped 2nd minus an original 4th have a difference of 4, but a flipped 4th minus an original 2nd have a difference of at least 8 (we still don't know if the thousands will be lending to the hundreds, yet) WITHOUT any borrowing, thus flipped 4th ≥ original 2nd + 8, while original 4th have the same parity with flipped 2nd. Therefore, this limited the flipped 4th to 8 or 9 and original 4th to 8 or 6, so anyway the shared parity is even, for 2–8 or 0–6 paired with 8–2 or 9–0 respectively, an instant shortlist here. }

Now it's become
78633 = 89x01 – 10x68
We know that the hundreds place borrowed from the thousands and lent to the tens, both, so rotated 3rd < original 3rd, and rotated + 10 – 1 = rotated + 9 = original + 6. It's impossible for the self-representing singles to be both when the two should have a difference of 3, therefore the 6–9 pair is the only possibility here.

Answer = 10968