Difference between revisions of "2019 AMC 8 Problems/Problem 25"

Latest revision as of 18:49, 6 August 2023

Problem 25

Alice has $24$ 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? $\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380$

Solution 1

We use stars and bars. Let Alice get $k$ apples, let Becky get $r$ apples, let Chris get $y$ apples. $\implies k + r + y = 24$ We can manipulate this into an equation which can be solved using stars and bars.

All of them get at least $2$ apples, so we can subtract $2$ from $k$ , $2$ from $r$ , and $2$ from $y$ . $\implies (k - 2) + (r - 2) + (y - 2) = 18$ Let $k' = k - 2$ , let $r' = r - 2$ , let $y' = y - 2$ . $\implies k' + r' + y' = 18$ We can allow either of them to equal to $0$ ; hence, this can be solved by stars and bars.

By Stars and Bars, our answer is just $\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}$ .

Solution 2 (Answer Choices)

Consider an unordered triple $(a,b,c)$ where $a+b+c=24$ and $a,b,c$ are not necessarily distinct. Then, we will either have $1$ , $3$ , or $6$ distinguishable ways to assign $a$ , $b$ , and $c$ to Alice, Becky, and Chris. Thus, our answer will be $x+3y+6z$ for some nonnegative integers $x,y,z$ . Notice that we only have $1$ way to assign the numbers $a,b,c$ to Alice, Becky, and Chris when $a=b=c$ . As this only happens $1$ way ( $a=b=c=8$ ), our answer is $1+3y+6z$ for some $y,z$ . Finally, notice that this implies the answer is $1$ mod $3$ . The only answer choice that satisfies this is $\boxed{\textbf{(C) }190}$ .

-BorealBear

Solution 3

Since each person needs to have at least two apples, we can simply give each person two, leaving $24 - 2\times3=18$ apples. For the remaining apples, if Alice is going to have $a$ apples, Becky is going to have $b$ apples, and Chris is going to have $c$ apples, we have indeterminate equation $a+b+c=18$ . Currently, we can see that $0 \leq a\leq 18$ where $a$ is an integer, and when $a$ equals any number in the range, there will be $18-a+1=19-a$ sets of values for $b$ and $c$ . Thus, there are $19 + 18 + 17 + \cdots + 1 = \boxed{\textbf{(C) }190}$ possible sets of values in total.

~Bloggish

Video Solution by OmegaLearn

https://youtu.be/5UojVH4Cqqs?t=5131

~ pi_is_3.14

Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)

https://youtu.be/Xm4ZGND9WoY

~Hayabusa1

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

https://youtu.be/8kzjB60pBrA

~savannahsolver

@@ Line 1: / Line 1: @@
 ==Problem 25==
 Alice has <math>24</math> 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?
+<math>\textbf{(A) }105\qquad\textbf{(B) }114\qquad\textbf{(C) }190\qquad\textbf{(D) }210\qquad\textbf{(E) }380</math>
 ==Solution 1==
-We use [[stars and bars]]. The problem asks for the number of integer solutions <math>(a,b,c)</math> such that <math>a+b+c = 24</math> and <math>a,b,c \ge 2</math>. We can subtract 2 from <math>a</math>, <math>b</math>, <math>c</math>, so that we equivalently seek the number of non-negative integer solutions to <math>a' + b' + c' = 18</math>. By stars and bars (using 18 stars and 2 bars), the number of solutions is <math>\binom{18+2}{2} = \binom{20}{2} = \boxed{\textbf{(C) }190}</math>.
+We use [[stars and bars]]. Let Alice get <math>k</math> apples, let Becky get <math>r</math> apples, let Chris get <math>y</math> apples.
+<cmath>\implies k + r + y = 24</cmath>We can manipulate this into an equation which can be solved using stars and bars.
-==Solution 2==
+All of them get at least <math>2</math> apples, so we can subtract <math>2</math> from <math>k</math>, <math>2</math> from <math>r</math>, and <math>2</math> from <math>y</math>.
-Let's say you assume that Alice has 2 apples. There are 19 ways to split the rest of the apples with Becky and Chris. If Alice has 3 apples, there are 18 ways to split the rest of the apples with Becky and Chris. If Alice has 4 apples, there are 17 ways to split the rest. So the total number of ways to split 24 apples between the three friends is equal to 19 + 18 + 17...…… + 1 = 20 (19/2) = <math>\boxed{\textbf{(C)}\ 190}</math>
+<cmath>\implies (k - 2) + (r - 2) + (y - 2) = 18</cmath>Let <math>k' = k - 2</math>, let <math>r' = r - 2</math>, let <math>y' = y - 2</math>.
+<cmath>\implies k' + r' + y' = 18</cmath>We can allow either of them to equal to <math>0</math>; hence, this can be solved by stars and bars.
-~heeeeeeheeeeeee
+By Stars and Bars, our answer is just <math>\binom{18 + 3 - 1}{3 - 1} = \binom{20}{2} = \boxed{\textbf{(C)}\ 190}</math>.
+==Solution 2 (Answer Choices)==
+Consider an unordered triple <math> (a,b,c) </math> where <math> a+b+c=24 </math> and <math> a,b,c </math> are not necessarily distinct. Then, we will either have <math> 1 </math>, <math> 3 </math>, or <math> 6 </math> distinguishable ways to assign <math> a </math>, <math> b </math>, and <math> c </math> to Alice, Becky, and Chris. Thus, our answer will be <math> x+3y+6z </math> for some nonnegative integers <math> x,y,z </math>. Notice that we only have <math> 1 </math> way to assign the numbers <math> a,b,c </math> to Alice, Becky, and Chris when <math> a=b=c </math>. As this only happens <math> 1 </math> way (<math>a=b=c=8</math>), our answer is <math> 1+3y+6z </math> for some <math> y,z </math>. Finally, notice that this implies the answer is <math> 1 </math> mod <math> 3 </math>. The only answer choice that satisfies this is <math> \boxed{\textbf{(C) }190} </math>.
+-BorealBear
+==Solution 3==
+Since each person needs to have at least two apples, we can simply give each person two, leaving <math> 24 - 2\times3=18 </math> apples. For the remaining apples, if Alice is going to have <math> a </math> apples, Becky is going to have <math> b </math> apples, and Chris is going to have <math> c </math> apples, we have indeterminate equation <math> a+b+c=18 </math>. Currently, we can see that <math> 0 \leq a\leq 18 </math> where <math> a </math> is an integer, and when <math> a </math> equals any number in the range, there will be <math> 18-a+1=19-a </math> sets of values for <math> b </math> and <math> c </math>. Thus, there are <math> 19 + 18 + 17 + \cdots + 1 = \boxed{\textbf{(C) }190} </math> possible sets of values in total.
+~[[User:Bloggish|Bloggish]]
+== Video Solution by OmegaLearn ==
+https://youtu.be/5UojVH4Cqqs?t=5131
+~ pi_is_3.14
+==Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)==
+https://youtu.be/Xm4ZGND9WoY
+~Hayabusa1
+==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
+https://youtu.be/8kzjB60pBrA
+~savannahsolver
 ==See Also==
-{{AMC8 box|year=2019|num-b=24|after=Last Problem}}
 {{MAA Notice}}
+[[Category:Introductory Combinatorics Problems]]

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