2023 AMC 10A Problem 9

The year $2023$ contributes two $2$s (already even), one $0$ and one $3$. So the four digits $MMDD$ must contain an odd number of $0$s, an odd number of $3$s, and an even number of every other digit.

Three $0$s with a $3$, or three $3$s with a $0$, cannot form a valid month-day. So $MMDD$ consists of exactly one $0$, one $3$, and a pair $d, d$ with $d \ne 0, 3$. The month uses two of these four digits.

• $d = 1$: digits $\{0, 3, 1, 1\}$. Month $01$: days $13, 31$. Month $10$: days $13, 31$. Month $11$: days $03, 30$. Month $03$: day $11$. That is $7$ dates.
• $d = 2$: digits $\{0, 3, 2, 2\}$. Month $02$: day $23$ (day $32$ invalid). Month $03$: day $22$. Month $12$ is not available since it would use a $1$. That is $2$ dates.
• $d \ge 4$: the month would have to be $0d$ or $d0$, needing a day made of $3$ and $d$; days $3d$ and $d3$ with $d \ge 4$ are all above $31$. No dates.

Total $7 + 2 = 9$.

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

Parity of digit counts is a multiset condition, so first translate the year's contribution into a requirement on the remaining four digits. Then split by the one free choice (the paired digit) and let the calendar constraints prune the list. Fixed-format enumeration problems reward writing the constraint before listing anything.

• Counting 1203 and 1230 in the pair-of-2s case even though month 12 contributes a 1, breaking the digit pattern; or missing 0311.
• Forgetting that the $0$ and $3$ can both land in the day ($1103$, $1130$) rather than one in each part.