Art of Problem Solving

2025 AMC 8 Problems/Problem 3: Difference between revisions

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~shreyan.chethan
~shreyan.chethan
==Vide Solution 1 by SpreadTheMathLove==
https://www.youtube.com/watch?v=jTTcscvcQmI


==Video Solution by Daily Dose of Math==
==Video Solution by Daily Dose of Math==
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~Thesmartgreekmathdude
~Thesmartgreekmathdude


==Vide Solution 1 by SpreadTheMathLove==
https://www.youtube.com/watch?v=jTTcscvcQmI
==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 17:49, 30 January 2025

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?


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

Vide Solution 1 by SpreadTheMathLove

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

Video Solution by Daily Dose of Math

https://youtu.be/rjd0gigUsd0

~Thesmartgreekmathdude

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
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

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