Difference between revisions of "2025 AMC 8 Problems/Problem 3"
(→Solution 2) |
|||
| Line 8: | Line 8: | ||
==Solution 2== | ==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 | + | 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== | ==Vide Solution 1 by SpreadTheMathLove== | ||
Revision as of 12:49, 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 
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
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. 
