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? | ''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> | <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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| − | We start with <math>4</math> players | + | We start with Annika and <math>3</math> of her friends playing, meaning that there are <math>4</math> players. This must mean that there is a total of <math>4 \cdot 15 = 60</math> cards. If <math>2</math> more players joined, there would be <math>6</math> players, and since the cards need to be split evenly, this would mean that each player gets <math>\frac{60}{6}=\boxed{\text{(C)\ 10}}</math> cards. |
~shreyan.chethan | ~shreyan.chethan | ||
| − | == Video Solution | + | == Video Solution 1 (Detailed Explanation) == |
| − | https://youtu.be/ | + | https://youtu.be/TbwqZy5_Q18 |
| + | ~ ChillGuyDoesMath | ||
| − | == Video Solution | + | == Video Solution 2 by SpreadTheMathLove == |
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| − | https:// | + | https://www.youtube.com/watch?v=jTTcscvcQmI |
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| − | ==Video Solution | + | == Video Solution 3 == |
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https://youtu.be/VP7g-s8akMY?si=ciMC0Hje4_kpkd3P&t=158 | https://youtu.be/VP7g-s8akMY?si=ciMC0Hje4_kpkd3P&t=158 | ||
~hsnacademy | ~hsnacademy | ||
| − | ==Video Solution by Daily Dose of Math== | + | ==Video Solution 4 by Daily Dose of Math== |
https://youtu.be/rjd0gigUsd0 | https://youtu.be/rjd0gigUsd0 | ||
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~Thesmartgreekmathdude | ~Thesmartgreekmathdude | ||
| − | ==Video Solution by Feetfinder== | + | == Video Solution 5 by Feetfinder == |
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https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | == Video Solution by | + | |
| + | == Video Solution 6 by CoolMathProblems== | ||
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https://youtu.be/tu0rZLUSQFg | https://youtu.be/tu0rZLUSQFg | ||
| − | ==See Also== | + | == Video Solution 7 by Pi Academy == |
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| + | https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK | ||
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| + | == 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}} | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
Latest revision as of 18:41, 3 February 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
We start with Annika and of her friends playing, meaning that there are 
players. This must mean that there is a total of 
cards. If more players joined, there would be 
players, and since the cards need to be split evenly, this would mean that each player gets 
cards.
~shreyan.chethan
Video Solution 1 (Detailed Explanation)
~ ChillGuyDoesMath
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
~Thesmartgreekmathdude
Video Solution 5 by Feetfinder
Video Solution 6 by CoolMathProblems
Video Solution 7 by Pi Academy
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
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. 
