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Difference between revisions of "2019 AMC 8 Problems/Problem 23"

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Starting from the above equation <math>\frac{t}{4}+\frac{2t}{7} + 15 + x = t</math> where <math>t</math> is the total number of points scored and <math>x\le 14</math> is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation <math>x+15 = \frac{13}{28}t</math>, or <math>28x+28\cdot 15=13t</math>. Since <math>t</math> is necessarily divisible by 28, let <math>t=28u</math> where <math>u \ge 0</math> and divide by 28 to obtain <math>x + 15 = 13u</math>. Then it is easy to see <math>u=2</math> (<math>t=56</math>) is the only candidate, giving <math>x=\boxed{\textbf{(B)} 11}</math>. -scrabbler94
 
Starting from the above equation <math>\frac{t}{4}+\frac{2t}{7} + 15 + x = t</math> where <math>t</math> is the total number of points scored and <math>x\le 14</math> is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation <math>x+15 = \frac{13}{28}t</math>, or <math>28x+28\cdot 15=13t</math>. Since <math>t</math> is necessarily divisible by 28, let <math>t=28u</math> where <math>u \ge 0</math> and divide by 28 to obtain <math>x + 15 = 13u</math>. Then it is easy to see <math>u=2</math> (<math>t=56</math>) is the only candidate, giving <math>x=\boxed{\textbf{(B)} 11}</math>. -scrabbler94
  
−
==Solution 3==
+
==Solution 2==
 
We first start by setting the total number of points as <math>28</math>, since <math>\text{LCM}(4,7) = 28</math>. However, we see that this does not work since we surpass the number of points just with the information given (<math>28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30</math> <math>(> 28)</math> ). Next, we can see that the total number of points scored is <math>56</math> as, if it is more than or equal to <math>84</math>, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: <math>56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45</math>, and thus, the other seven players would have scored a total of <math>56-45 = \boxed{\textbf{(B)} 11}</math> (We see that this works since we could have <math>4</math> of them score <math>2</math> points, and the other <math>3</math> of them score <math>1</math> point) -aops5234 -Edited by [[User: Penguin_Spellcaster|Penguin_Spellcaster]]
 
We first start by setting the total number of points as <math>28</math>, since <math>\text{LCM}(4,7) = 28</math>. However, we see that this does not work since we surpass the number of points just with the information given (<math>28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30</math> <math>(> 28)</math> ). Next, we can see that the total number of points scored is <math>56</math> as, if it is more than or equal to <math>84</math>, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: <math>56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45</math>, and thus, the other seven players would have scored a total of <math>56-45 = \boxed{\textbf{(B)} 11}</math> (We see that this works since we could have <math>4</math> of them score <math>2</math> points, and the other <math>3</math> of them score <math>1</math> point) -aops5234 -Edited by [[User: Penguin_Spellcaster|Penguin_Spellcaster]]
  

Revision as of 10:12, 30 April 2021

Problem 23

After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?

$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14$

Solution 1

Starting from the above equation $\frac{t}{4}+\frac{2t}{7} + 15 + x = t$ where $t$ is the total number of points scored and $x\le 14$ is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation $x+15 = \frac{13}{28}t$, or $28x+28\cdot 15=13t$. Since $t$ is necessarily divisible by 28, let $t=28u$ where $u \ge 0$ and divide by 28 to obtain $x + 15 = 13u$. Then it is easy to see $u=2$ ($t=56$) is the only candidate, giving $x=\boxed{\textbf{(B)} 11}$. -scrabbler94

Solution 2

We first start by setting the total number of points as $28$, since $\text{LCM}(4,7) = 28$. However, we see that this does not work since we surpass the number of points just with the information given ($28\cdot\frac{1}{4}+28\cdot\frac{2}{7} + 15 = 30$ $(> 28)$ ). Next, we can see that the total number of points scored is $56$ as, if it is more than or equal to $84$, at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: $56\cdot\frac{1}{4}+56\cdot\frac{2}{7} + 15 = 45$, and thus, the other seven players would have scored a total of $56-45 = \boxed{\textbf{(B)} 11}$ (We see that this works since we could have $4$ of them score $2$ points, and the other $3$ of them score $1$ point) -aops5234 -Edited by Penguin_Spellcaster

Solution 4 — Modular Arithmetic

Adding together Alexa's and Brittany's fractions, we get $\frac{15}{28}$ as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: $\frac{15x}{28x}$ where $x$ is the common ratio. Let $y$ and $z$ and $w$ be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have $\frac{13x}{28x} = 15 + y + 2z + 0w.$ We want all of our variables to be integers. Thus, we want $15 + y + 2z = 0 \pmod {13}.$ Simplifying, $y+2z = 11 \pmod {13}.$ The only possible value, as this integer sum has to be less than $7 \cdot 2 + 1 = 15,$ must be 11. Therefore $y+2z = 11,$ and the answer is $\boxed{ \textbf{(B) 11}}$ - ab2024

Video explaining solution

https://www.youtube.com/watch?v=fKjmw_zzCUU - Happytwin

Associated video - https://www.youtube.com/watch?v=jE-7Se7ay1c

https://www.youtube.com/watch?v=3Mae_6qFxoU&t=204s ~ hi_im_bob

https://youtu.be/wsYCn2FqZJE

https://www.youtube.com/watch?v=o2mcnLOVFBA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=5 ~ MathEx

Video Solution

https://youtu.be/HISL2-N5NVg?t=4115

~ pi_is_3.14

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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

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