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2025 AMC 8 Problems/Problem 3: Difference between revisions

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==Problem==
== Problem ==
 
''Buffalo Shuffle-o'' is a card game in which all the cards are distributed evenly among all players at the start of the game. When Annika and <math>3</math> of her friends play ''Buffalo Shuffle-o'', each player is dealt <math>15</math> cards. Suppose <math>2</math> more friends join the next game. How many cards will be dealt to each player?
''Buffalo Shuffle-o'' is a card game in which all the cards are distributed evenly among all players at the start of the game. When Annika and <math>3</math> of her friends play ''Buffalo Shuffle-o'', each player is dealt <math>15</math> cards. Suppose <math>2</math> more friends join the next game. How many cards will be dealt to each player?


<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12</math>
<math>\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12</math>


==Solution 1==
== Solution 1 ==
 
In the beginning, there are <math>4</math> players playing the game, meaning that there is a total of <math>4 \cdot 15 = 60</math> cards. When <math>2</math> more players join, there are now <math>6</math> players playing, and since the cards need to be split evenly, this means that each player gets <math>\frac{60}{6}=\boxed{\text{(C)\ 10}}</math> cards
 
==Solution 2==


We start with <math>4</math> players playing the game (Anika + <math>3</math> friends). When <math>2</math> more players join, there are now <math>6</math> players playing, and since the cards need to be split evenly, this means we can set up the equation <math>\frac{4 \cdot 15}{6} = \frac{60}{6}=\boxed{\text{(C)\ 10}}</math> cards  
We start with Annika and <math>3</math> of her friends playing, meaning that there are <math>4</math> players. This must mean that there is a total of <math>4 \cdot 15 = 60</math> cards. If <math>2</math> more players joined, there would be <math>6</math> players, and since the cards need to be split evenly, this would mean that each player gets <math>\frac{60}{6}=\boxed{\text{(C)\ 10}}</math> cards.


~shreyan.chethan
~shreyan.chethan


== Video Solution by Pi Academy ==
== Video Solution 1 (Detailed Explanation) ==


https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
https://youtu.be/TbwqZy5_Q18


~ ChillGuyDoesMath


== Video Solution 1 (Detailed Explanation) 🚀⚡📊 ==
== Video Solution 2 by SpreadTheMathLove ==
Youtube Link ⬇️


https://youtu.be/TbwqZy5_Q18
https://www.youtube.com/watch?v=jTTcscvcQmI
 
~ ChillGuyDoesMath :)


==Video Solution by SpreadTheMathLove==
== Video Solution 3 ==
https://www.youtube.com/watch?v=jTTcscvcQmI


==Video Solution (A Clever Explanation You’ll Get Instantly)==
https://youtu.be/VP7g-s8akMY?si=ciMC0Hje4_kpkd3P&t=158
https://youtu.be/VP7g-s8akMY?si=ciMC0Hje4_kpkd3P&t=158
~hsnacademy
~hsnacademy


==Video Solution by Daily Dose of Math==
==Video Solution 4 by Daily Dose of Math==


https://youtu.be/rjd0gigUsd0
https://youtu.be/rjd0gigUsd0
Line 39: Line 32:
~Thesmartgreekmathdude
~Thesmartgreekmathdude


==Video Solution by Feetfinder==
== Video Solution 5 by Feetfinder ==
 
https://youtu.be/PKMpTS6b988
https://youtu.be/PKMpTS6b988
== Video Solution by CoolMathProblem s==
 
== Video Solution 6 by CoolMathProblems==
 
https://youtu.be/tu0rZLUSQFg
https://youtu.be/tu0rZLUSQFg


==See Also==
== Video Solution 7 by Pi Academy ==
 
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
 
== See Also ==
 
{{AMC8 box|year=2025|num-b=2|num-a=4}}
{{AMC8 box|year=2025|num-b=2|num-a=4}}
{{MAA Notice}}
{{MAA Notice}}
[[Category:Introductory Number Theory Problems]]

Revision as of 18:41, 3 February 2025

Problem

Buffalo Shuffle-o is a card game in which all the cards are distributed evenly among all players at the start of the game. When Annika and $3$ of her friends play Buffalo Shuffle-o, each player is dealt $15$ cards. Suppose $2$ more friends join the next game. How many cards will be dealt to each player?

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 11 \qquad \textbf{(E)}\ 12$

Solution 1

We start with Annika and $3$ of her friends playing, meaning that there are $4$ players. This must mean that there is a total of $4 \cdot 15 = 60$ cards. If $2$ more players joined, there would be $6$ players, and since the cards need to be split evenly, this would mean that each player gets $\frac{60}{6}=\boxed{\text{(C)\ 10}}$ cards.

~shreyan.chethan

Video Solution 1 (Detailed Explanation)

https://youtu.be/TbwqZy5_Q18

~ ChillGuyDoesMath

Video Solution 2 by SpreadTheMathLove

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

Video Solution 3

https://youtu.be/VP7g-s8akMY?si=ciMC0Hje4_kpkd3P&t=158 ~hsnacademy

Video Solution 4 by Daily Dose of Math

https://youtu.be/rjd0gigUsd0

~Thesmartgreekmathdude

Video Solution 5 by Feetfinder

https://youtu.be/PKMpTS6b988

Video Solution 6 by CoolMathProblems

https://youtu.be/tu0rZLUSQFg

Video Solution 7 by Pi Academy

https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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