2019 AMC 8 Problems/Problem 25
Contents
Problem 25
Alice has apples. In how many ways can she share them with Becky and Chris so that each of the three people has at least two apples?
Solution 1
We use stars and bars. Let Alice get apples, let Becky get apples, let Chris get apples. We can manipulate this into an equation which can be solved using stars and bars.
All of them get at least apples, so we can subtract from , from , and from . Let , let , let . We can allow either of them to equal to ; hence, this can be solved by stars and bars.
By Stars and Bars, our answer is just .
Solution 2
First assume that Alice has apples. There are ways to split the rest of the apples with Becky and Chris. If Alice has apples, there are ways to split the rest of the apples with Becky and Chris. If Alice has apples, there are ways to split the rest. So, the total number of ways to split apples between the three friends is equal to .
Solution (Answer Choices)
Consider an unordered triple where and are not necessarily distinct. Then, we will either have , , or distinguishable ways to assign , , and to Alice, Becky, and Chris. Thus, our answer will be for some nonnegative integers . Notice that we only have way to assign the numbers to Alice, Becky, and Chris when . As this only happens way (), our answer is for some . Finally, notice that this implies the answer is mod . The only answer choice that satisfies this is .
-BorealBear
Video Solution by OmegaLearn
https://youtu.be/5UojVH4Cqqs?t=5131
~ pi_is_3.14
Video Solutions
https://www.youtube.com/watch?v=EJzSOPXULBc
- Happytwin
https://www.youtube.com/watch?v=wJ7uvypbB28
https://www.youtube.com/watch?v=2dBUklyUaNI
https://www.youtube.com/watch?v=3qp0wTq-LI0&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=7
~ MathEx
~savannahsolver
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
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