2025 AMC 10A Problems/Problem 12

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?

$\textbf{(A) } 176 \qquad\textbf{(B) } 192 \qquad\textbf{(C) } 432 \qquad\textbf{(D) } 464 \qquad\textbf{(E) } 608$

Solution 1: Casework

The only two digits that aren't prime and aren't even are $1$ and $9.$ We split this problem into cases on the number of $2$ s (since $2$ is both a prime number and an even number).

Case $1:$ No $2$ s. For this case, there are $4$ choices for where the even digit goes, and $3$ choices for what it is. There are then $3$ choices for where the prime digit goes, and $3$ choices for what it is. The last two spots have $2$ choices each, $1$ or $9.$ This gives a total of $4\cdot 3\cdot 3\cdot 3\cdot 2^2 = 432$ options for this case.

Case $2:$ One $2$ . For this case, there are $4$ choices for where $2$ goes, and $2$ choices for the other three digits each. This case gives a total of $2^3\cdot 4 = 32$ options. Hence, the answer is $432 + 32 = \boxed{\text{(D) }464}.$

~Tacos_are_yummy_1

~iiiiiizh (minor edits)

~drekie (very minor edit--ain't no way someone thought 4x3x3x2x2=432 lmao)

~happyfish0922 (minor formatting edits)

~zoyashaikh (extremely minor edits)

Video Solution (Fast and Easy)

https://youtu.be/TOTJEltmpe0?si=pglACCfjzAHguvlt ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

2025 AMC 10A (Problems • Answer Key • Resources)
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2025 AMC 10A Problems/Problem 12

Contents

Solution 1: Casework

Video Solution (Fast and Easy)

Video Solution

See Also