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

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==Video Solution by Feetfinder==
 
==Video Solution by Feetfinder==
 
https://youtu.be/PKMpTS6b988
 
https://youtu.be/PKMpTS6b988
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== Video Solution by CoolMathProblem s==
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https://youtu.be/tu0rZLUSQFg
  
 
==See Also==
 
==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}}

Revision as of 23:56, 1 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

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

Solution 2

We start with $4$ players playing the game (Anika + $3$ friends). When $2$ more players join, there are now $6$ players playing, and since the cards need to be split evenly, this means we can set up the equation $\frac{4 \cdot 15}{6} = \frac{60}{6}=\boxed{\text{(C)\ 10}}$ cards

~shreyan.chethan

Video Solution 1 (Detailed Explanation) 🚀⚡📊

Youtube Link ⬇️

https://youtu.be/TbwqZy5_Q18

~ ChillGuyDoesMath :)

Video Solution by SpreadTheMathLove

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 ~hsnacademy

Video Solution by Daily Dose of Math

https://youtu.be/rjd0gigUsd0

~Thesmartgreekmathdude

Video Solution by Feetfinder

https://youtu.be/PKMpTS6b988

Video Solution by CoolMathProblem s

https://youtu.be/tu0rZLUSQFg

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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