Difference between revisions of "2019 AMC 8 Problems/Problem 23"
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==Solution 3== | ==Solution 3== | ||
− | 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 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]] |
==Video explaining solution== | ==Video explaining solution== |
Revision as of 03:20, 1 November 2020
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
Since and are integers, we have . We see that the number of points scored by the other team members is less than or equal to and greater than or equal to . We let the total number of points be and the total number of points scored by the other team members be , which means that , which means . The only value of that satisfies all conditions listed is , so . - juliankuang (lol im smart)
Solution 2
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 3
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
Video explaining solution
https://www.youtube.com/watch?v=fKjmw_zzCUU
https://www.youtube.com/watch?v=o2mcnLOVFBA&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=5 ~ MathEx
https://www.youtube.com/watch?v=aStuVhoD8wc- Also includes other problems from 21-25
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.