Transforming a 4-digit number to produce $7793$

Question

Harvey transforms a four-digit number by reversing the order of its digits, subtracting 1 from all digits that are 1 more than a multiple of 3, and adding 1 to all even digits in that order. Harvey obtains 7793 after transformation. Find the number of distinct numbers that he could have transformed. (MAT 2021 Sample Question 2)

I first noted that since all the digits of $7793$ are odd, Harvey must have added $1$ all of the digits, so the number prior to the last transformation was $6682.$ Then, since the second transformation must have turned each digit into a multiple of three, I added $1$ to each $6$ to give me $7782.$ Then, I reversed the digits to get $2877,$ which appears to be the only solution.

However, the sample solutions state that the answer is actually $24.$ Can somebody explain where I messed up?

Answer 1

A digit $7$ in the final number may be the result of adding one to the even digit $6$, or by leaving the odd digit $7$ as is.

If we ignore the first step (digit reversal), which is reversible, you may perhaps check what a digit $0$, digit $1$, ... , digit $9$ turn into after the second and third step - which of these become $7$ or $9$ or $3$?

Answer 2

While it states that:

"Harvey obtains $7793$ after transformation"

all we know is that Harvey added $1$ to all the even digits of what he had before. That does not mean "all the digits of that number were even".

Hence he could have obtained, say, $7792$ as a result of his previous operation.