Difference between revisions of "2025 AMC 8 Problems/Problem 3"
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| + | ==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> | ||
==Solution 1== | ==Solution 1== | ||
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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 | 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 | ||
| + | |||
| + | ~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 | ||
| + | |||
| + | ==Video Solution by Thinking Feet== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=2|num-a=4}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:17, 30 January 2025
Contents
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 
of her friends play Buffalo Shuffle-o, each player is dealt 
cards. Suppose 
more friends join the next game. How many cards will be dealt to each player?
Solution 1
In the beginning, there are 
players playing the game, meaning that there is a total of 
cards. When more players join, there are now 
players playing, and since the cards need to be split evenly, this means that each player gets 
cards
Solution 2
We start with players playing the game (Anika + friends). When more players join, there are now players playing, and since the cards need to be split evenly, this means we can set up the equation 
cards
~shreyan.chethan
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
Video Solution by Thinking Feet
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. 
