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2025 AMC 10A Problems/Problem 12

Problem

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 are neither prime nor even are $1$ and $9.$ We split this problem into cases based on the number of $2$s. This is because $2$ is both a prime number and an even number.

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

Case 2: For this case, there is one $2$. 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 $432 + 32 = \boxed{\textbf{(D) }464}$ ~Tacos_are_yummy_1

~Tacos_are_yummy_1 (w editing chain, let's keep it going haha)

~iiiiiizh (minor edits)

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

~happyfish0922 (minor formatting edits)

~zoyashaikh (extremely minor edits)

~kfclover (minor LaTeX edits)

~aldzandrtc (removed dummy subjects to be specific)

~gvh300 (minor editing)

~minor $\LaTeX$ edits by i_am_not_suk_at_math (saharshdevaraju 14:27, 7 November 2025 (EST)saharshdevaraju)

~Ninekayx wsp

~Strickenox (added an exclamation point just to keep the chain going)

~SixthGradeBookWorm927 (Made the first sentence for each case complete)

~ Strickenox (wsp SixthGradeBookworm927 i see we are both editing this page right now)

~Sharp_logic (made first sentence complete)

~Harrg (very minor edits)

~ZDC0530 (minor LaTeX edit)

~HappyLion (extremely minor grammar mistake)

~Elliecorn (very extremely minor LaTeX edit)

~EZ123 (extremely minor formatting edit to make the solution more pleasing to the eye)

~Galactic_Saber (extremely minor formatting edit)

~lucassf12 (removed period at the end to keep the chain going)

~Ryxo (minor grammar improvements for better flow)

~AoPS_enjoyer (removed unnecessary parenthesis)

~piZZaZedpiZZa (removed space before the colon)

Solution 2: Cheese but fast

Let us count the cases where there is a $2$ anywhere in the lock. There are $4$ places to arrange the $2$s, and the remaining digits can only be $1$s and $9$s (since they neither can be even nor prime). Thus, there will be $4\cdot2^3=32$ choices for there to exist one $2$. We note that answer choice $\text{(D)}$ is $32$ greater than another answer choice (specifically, $\text{(C)}$) and is the only answer choice to have this property. Assuming that the pitfall of forgetting the case with a $2$ would be a separate answer choice on its own, we can justify that the answer is $\boxed{\text{(D) }464}$

~megaboy6679, who forgot the case with the $2$

Video Solution

https://youtu.be/CCYoHk2Af34

Chinese Video Solution

https://www.bilibili.com/video/BV1nYkUBVEFt/

~metrixgo

Video Solution (Fast and Easy)

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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

Video Solution by Daily Dose of Math

https://youtu.be/5Fjos1vBt0A

~Thesmartgreekmathdude

See Also

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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