2025 AMC 10A Problem 12

P11-15~5 minPrint

Basic Counting Arrangements with Restrictions Split into exhaustive cases and handle each Set up the equation/formula and compute; no special trick needed

Carlos uses a $4$ -digit passcode to unlock his computer. In his passcode, exactly one digit is even, exactly one (possibly different) digit is prime, and no digit is $0$ . How many $4$ -digit passcodes satisfy these conditions?

Show solution without answering

Key insight

Sort digits 1 to 9 into groups: 2 is even and prime, 4, 6, 8 only even, 3, 5, 7 only prime, 1 and 9 neither. Then count two cases.

Solution

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

Sort the digits.

Group	Digits	How many
even and prime	$2$	$1$
even, not prime	$4, 6, 8$	$3$
prime, not even	$3, 5, 7$	$3$
neither	$1, 9$	$2$

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

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

Why this works

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.5)
Introduction to Counting & Probability, Chapter 3: Correcting for Overcounting (§3.2, §3.4)
Prealgebra, Chapter 14: Counting (§14.3)