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Difference between revisions of "2025 AMC 8 Problems/Problem 3"

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== Problem ==
 
== Problem ==
  
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''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?
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''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>67</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>

Revision as of 22:59, 2 November 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 $67$ 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 $\frac{60}{6}=\boxed{\text{(C)\ 10}}$ Buffalo Shuffle-O cards each meaning that the final answer is 10.

~shreyan.chethan

Video Solution 1 (Detailed Explanation) 🚀⚡📊

https://www.youtube.com/watch?v=TbwqZy5_Q18

~ ChillThingz :) 656565765675675

Video Solution! Super simple and insanely fast!

https://youtu.be/1NQQ7dsHHKE

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

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

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. AMC Logo.png

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