2025 AMC 10A Problem 12

Two rules overlap here (one even digit, one prime digit), so first sort the digits $1$ to $9$ by which rules they touch.

1. Sort the digits.

Remember that $1$ is not prime, and $9 = 3 \cdot 3$ is not prime.

1. Split into two cases. The even digit and the prime digit might be the same digit, or two different digits.

3. Case 1: one digit is both even and prime. That digit must be $2$. The other three digits can be neither even nor prime, so each is $1$ or $9$.
- Where does the $2$ go? $4$ choices.
- Each of the other three spots: $2$ choices.
$$4 \cdot 2 \cdot 2 \cdot 2 = 32.$$

4. Case 2: two different digits. One spot has an even digit that is not prime, and another spot has a prime digit that is not even. A $2$ cannot appear here, since it would add a second even digit and a second prime digit. The last two spots are $1$ or $9$.
- Spot for the even digit: $4$ choices.
- Spot for the prime digit: $3$ choices (any other spot).
- Which even digit ($4$, $6$ or $8$): $3$ choices.
- Which prime digit ($3$, $5$ or $7$): $3$ choices.
- The two remaining spots: $2 \cdot 2 = 4$ choices.
$$4 \cdot 3 \cdot 3 \cdot 3 \cdot 4 = 432.$$

5. Add the cases. A passcode either has a $2$ (Case 1) or does not (Case 2), so no passcode is counted twice:
$$32 + 432 = 464.$$

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

When two conditions overlap, a small table of which items satisfy which condition shows every situation at once. Here it makes clear that $2$ is special and that only $1$ and $9$ can fill the leftover spots. After that, each case is just "choose the spots, then choose the digits" and multiply.

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 $\boxed{\textbf{(D)}\ 464}$.