Difference between revisions of "2019 AMC 8 Problems/Problem 23"
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==Problem 23== | ==Problem 23== | ||
− | After Euclid High School's last basketball game, it was determined that <math>\frac{1}{4}</math> of the team's points were scored by Alexa and <math>\frac{2}{7}</math> were scored by Brittany. Chelsea scored <math>15</math> points. None of the other <math>7</math> team members scored more than <math>2</math> points What was the total number of points scored by the other <math>7</math> team members? | + | After Euclid High School's last basketball game, it was determined that <math>\frac{1}{4}</math> of the team's points were scored by Alexa and <math>\frac{2}{7}</math> were scored by Brittany. Chelsea scored <math>15</math> points. None of the other <math>7</math> team members scored more than <math>2</math> points. What was the total number of points scored by the other <math>7</math> team members? |
<math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math> | <math>\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14</math> | ||
==Solution 1== | ==Solution 1== | ||
− | + | 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 2== | ==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]] | |
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+ | ==Solution 3 — Modular Arithmetic == | ||
+ | |||
+ | Adding together Alexa's and Brittany's fractions, we get <math>\frac{15}{28}</math> as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: <math>\frac{15x}{28x}</math> where <math>x</math> is the common ratio. Let <math>y</math> and <math>z</math> and <math>w</math> be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have <math>\frac{13x}{28x} = 15 + y + 2z + 0w.</math> We want all of our variables to be integers. Thus, we want <math>15 + y + 2z = 0 \pmod {13}.</math> Simplifying, <math>y+2z = 11 \pmod {13}.</math> The only possible value, as this integer sum has to be less than <math>7 \cdot 2 + 1 = 15,</math> must be 11. Therefore <math>y+2z = 11,</math> and the answer is <math>\boxed{ \textbf{(B) 11}}</math> - ab2024 | ||
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+ | ==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 | ||
+ | |||
+ | https://youtu.be/HISL2-N5NVg?t=4115 | ||
− | + | ~ pi_is_3.14 | |
− | |||
==See Also== | ==See Also== |
Latest revision as of 21:23, 9 July 2021
Contents
Problem 23
After Euclid High School's last basketball game, it was determined that of the team's points were scored by Alexa and were scored by Brittany. Chelsea scored points. None of the other team members scored more than points. What was the total number of points scored by the other team members?
Solution 1
Starting from the above equation where is the total number of points scored and is the number of points scored by the remaining 7 team members, we can simplify to obtain the Diophantine equation , or . Since is necessarily divisible by 28, let where and divide by 28 to obtain . Then it is easy to see () is the only candidate, giving . -scrabbler94
Solution 2
We first start by setting the total number of points as , since . However, we see that this does not work since we surpass the number of points just with the information given ( ). Next, we can see that the total number of points scored is as, if it is more than or equal to , at least one of the others will score more than 2 points. With this, we have that Alexa, Brittany, and Chelsea score: , and thus, the other seven players would have scored a total of (We see that this works since we could have of them score points, and the other of them score point) -aops5234 -Edited by Penguin_Spellcaster
Solution 3 — Modular Arithmetic
Adding together Alexa's and Brittany's fractions, we get as the fraction of the total number of points they scored together. However, this is just a ratio, so we can introduce a variable: where is the common ratio. Let and and be the number of people who scored 1, 2, and 0 points, respectively. Writing an equation, we have We want all of our variables to be integers. Thus, we want Simplifying, The only possible value, as this integer sum has to be less than must be 11. Therefore and the answer is - 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://www.youtube.com/watch?v=o2mcnLOVFBA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=5 ~ MathEx
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.