Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Video Solutions)
m (Reove duplicates)
Line 17: Line 17:
 
First assume that Alice has <math>2</math> apples. There are <math>19</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>3</math> apples, there are <math>18</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>4</math> apples, there are <math>17</math> ways to split the rest. So, the total number of ways to split <math>24</math> apples between the three friends is equal to <math>19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}</math>.
 
First assume that Alice has <math>2</math> apples. There are <math>19</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>3</math> apples, there are <math>18</math> ways to split the rest of the apples with Becky and Chris. If Alice has <math>4</math> apples, there are <math>17</math> ways to split the rest. So, the total number of ways to split <math>24</math> apples between the three friends is equal to <math>19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}</math>.
  
−
==Solution 3==
 
−
Let's assume that the three of them have <math>x, y, z</math> apples. Since each of them has to have at least <math>2</math> apples, we say that <math>a+2=x, b+2=y</math> and <math>c+2=z</math>. Thus, <math>a+b+c+6=24 \implies a+b+c=18</math>, and so by stars and bars, the number of solutions for this is <math>{n+k-1 \choose k-1} \implies {18+3-1 \choose 3-1} \implies {20 \choose 2}  = \boxed{\textbf{(C)}\ 190}</math>.
 
  
−
- aops5234
+
==Solution (Answer Choices)==
−
 
 
−
==Solution 4==
 
−
 
 
−
Since we have to give each of the <math>3</math> friends at least <math>2</math> apples, we need to spend a total of <math>2+2+2=6</math> apples to solve the restriction. Now we have <math>24-6=18</math> apples left to be divided among Alice, Becky, and Chris, without any constraints. We use the [[Ball-and-urn]] technique, or sometimes known as ([Sticks and Stones]/[Stars and Bars]), to divide the apples. We now have <math>18</math> stones and <math>2</math> sticks, which have a total of <math>\binom{18+2}{2}=\binom{20}{2}=\frac{20\times19}{2} = \boxed{190}</math> ways to arrange.
 
−
 
 
−
~by sakshamsethi
 
−
 
 
−
==Solution 5==
 
−
 
 
−
Equivalently, we split <math>21</math> apples among <math>3</math> friends with each having at least <math>1</math> apples. We put sticks between apples to split apples into three stacks. So there are 20 spaces to put <math>2</math> sticks. We have <math> \binom{20}{2} = 190</math> different ways to arrange the two sticks. So, there are <math>\boxed{190}</math> ways to split the apples among them.
 
−
 
 
−
~by Dolphindesigner
 
−
 
 
−
==Solution 6 (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> 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>.  
 
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> 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>.  
  

Revision as of 20:34, 3 January 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

First 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 $19 + 18 + 17...…… + 1 = 20\times \frac{19}{2}=\boxed{\textbf{(C)}\ 190}$.


Solution (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 $6$ 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

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/5UojVH4Cqqs?t=5131

~ pi_is_3.14

https://youtu.be/8kzjB60pBrA

~savannahsolver

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Problem
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_25&oldid=185657"