2025 AMC 10A Problem 12

The available digits are $1$ through $9$. Sort them by the two properties:

• even and prime: $2$ (one digit),
• even but not prime: $4, 6, 8$ (three digits),
• prime but not even: $3, 5, 7$ (three digits),
• neither: $1, 9$ (two digits).

Exactly one position holds an even digit and exactly one position holds a prime digit. Those two positions may or may not coincide.

One digit does both jobs. That digit must be $2$, and the other three digits must be neither even nor prime, so each is $1$ or $9$. Choose the position of the $2$ in $4$ ways and fill the rest in $2^3$ ways:
$$4 \cdot 2^3 = 32.$$

Two different positions. One position holds an even non-prime digit ($3$ choices) and a different position holds a prime non-even digit ($3$ choices); the other two positions are $1$ or $9$. Note that $2$ cannot appear here, since it would make the even count and the prime count both too large. Choosing the two special positions in order gives $4 \cdot 3$ ways:
$$(4 \cdot 3) \cdot 3 \cdot 3 \cdot 2^2 = 12 \cdot 36 = 432.$$

The two cases are disjoint, so the total is
$$32 + 432 = 464.$$

The answer is $\boxed{\textbf{(D)}\ 464}$.

Two overlapping conditions ("exactly one even" and "exactly one prime") are cleanest to handle by first classifying the alphabet of digits by which conditions each digit satisfies. Once the four classes are written down, the only structural question left is whether the even slot and the prime slot are the same slot, and each case is a straight multiplication. The classification also makes clear why $1$ and $9$ are the only legal fillers for the remaining positions, which is the step most easily lost when counting on the fly.

Choose the multiset of digit types first. If the passcode contains a $2$, the rest are from $\{1,9\}$: $4 \cdot 2^3 = 32$. If not, it contains exactly one of $\{4,6,8\}$ and exactly one of $\{3,5,7\}$ and two from $\{1,9\}$: choose which two of the four positions are special and which is which ($4 \cdot 3 = 12$), then the digits ($3 \cdot 3 \cdot 2 \cdot 2 = 36$). Same $32 + 432$.

• Counting only the case of two different special digits, 43 positions times 33 digits times 2^2, which gives 432, choice (C).
• Treating $2$ as merely even and allowing it alongside a separate prime digit, which double counts the prime condition.
• Allowing $1$ to count as prime, or forgetting that $9$ is neither even nor prime, so that the filler digits are miscounted.