2023 AMC 10A Problem 9

1. See what the year gives. The display starts with $2023$. That has two $2$s (already an even number), one $0$ and one $3$.
2. Decide what the month and day need. The four digits of the month and day must supply an odd number of $0$s, an odd number of $3$s, and an even number of every other digit.
3. Narrow it down. With four digits, the options are one $0$ and one $3$ plus a pair of another digit, or three of one and one of the other. Three $0$s and a $3$ (like $00\,03$) or three $3$s and a $0$ cannot be a real date. So the month and day use exactly one $0$, one $3$, and a pair $d, d$ of some other digit.
4. Case $d = 1$: the digits are $0, 3, 1, 1$. Go through the months that can be made from them:
- Month $01$ (January): days $13$ and $31$.
- Month $10$ (October): days $13$ and $31$.
- Month $11$ (November): days $03$ and $30$.
- Month $03$ (March): day $11$.

That is $7$ dates.
5. Case $d = 2$: the digits are $0, 3, 2, 2$.
- Month $02$ (February): day $23$. (Day $32$ does not exist.)
- Month $03$ (March): day $22$.
- Month $12$ is not possible, because it needs a $1$.

That is $2$ dates.
6. Case $d$ is $4$ or more: no month can use two copies of $d$ ($44$, $55$, and so on are not months), so the month must be $0d$ or $d0$, and only $0d$ is real. Then the day is made of $3$ and $d$: either $3d$ or $d3$, and both are more than $31$. No dates.
7. Total: $7 + 2 = 9$.

The answer is $\boxed{\textbf{(E)}\ 9}$.

"Each digit appears an even number of times" is about which digits are present, not their order. So first work out what the fixed part (the year) already supplies, and turn that into a rule for the remaining four digits. Then split into cases by the one free choice and let the calendar throw out impossible months and days.