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2025 AMC 10A Problems/Problem 12: Difference between revisions

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Carlos uses a <imath>4</imath>-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 <imath>0</imath>. How many <imath>4</imath>-digit passcodes satisfy these conditions?
==Problem==
Carlos uses a <imath>4</imath>-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 <imath>0</imath>. How many <imath>4</imath>-digit passcodes satisfy these conditions?


<imath>\textbf{(A) } 176 \qquad\textbf{(B) } 192 \qquad\textbf{(C) } 432 \qquad\textbf{(D) } 464 \qquad\textbf{(E) } 608</imath>
<imath>\textbf{(A) } 176 \qquad\textbf{(B) } 192 \qquad\textbf{(C) } 432 \qquad\textbf{(D) } 464 \qquad\textbf{(E) } 608</imath>


==Solution 1: Casework==
==Solution 1: Casework==
The only two digits that aren't prime and aren't even are <imath>1</imath> and <imath>9.</imath> We split this problem into cases on the number of <imath>2</imath>s (since <imath>2</imath> is both a prime number and an even number).
The only two digits that are neither prime nor even are <imath>1</imath> and <imath>9.</imath> We split this problem into cases based on the number of <imath>2</imath>s. This is because <imath>2</imath> is both a prime number and an even number.  


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


Case 2:  For this case, there is one <imath>2</imath>. There are <imath>4</imath> choices for where <imath>2</imath> goes, and <imath>2</imath> choices for the other three digits each. This case gives a total of <imath>2^3\cdot 4 = 32</imath> options!


Case <imath>2:</imath> One <imath>2</imath>. For this case, there are <imath>4</imath> choices for where <imath>2</imath> goes, and <imath>2</imath> choices for the other three digits each. This case gives a total of <imath>2^3\cdot 4 = 32</imath> options.
Hence, the answer is <imath>432 + 32 = \boxed{\textbf{(D) }464} </imath> ~Tacos_are_yummy_1
Hence, the answer is <imath>432 + 32 = \boxed{\textbf{(D) }464}. </imath>


~Tacos_are_yummy_1 (chat is this an editing chain?)
~Tacos_are_yummy_1 (w editing chain, let's keep it going haha)


~iiiiiizh (minor edits)
~iiiiiizh (minor edits)
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~kfclover (minor LaTeX edits)
~kfclover (minor LaTeX edits)


 
~aldzandrtc (removed dummy subjects to be specific)


~gvh300 (minor editing)
~gvh300 (minor editing)
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~minor <imath>\LaTeX</imath> edits by i_am_not_suk_at_math (saharshdevaraju 14:27, 7 November 2025 (EST)saharshdevaraju)
~minor <imath>\LaTeX</imath> edits by i_am_not_suk_at_math (saharshdevaraju 14:27, 7 November 2025 (EST)saharshdevaraju)


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


~megaboy6679, who forgot the case with the <imath>2</imath>
~megaboy6679, who forgot the case with the <imath>2</imath>
==Video Solution==
https://youtu.be/CCYoHk2Af34
==Chinese Video Solution==
https://www.bilibili.com/video/BV1nYkUBVEFt/
~metrixgo


== Video Solution (Fast and Easy) ==
== Video Solution (Fast and Easy) ==

Latest revision as of 20:58, 11 November 2025

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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