Difference between revisions of "2019 AMC 8 Problems/Problem 23"
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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? | ||
Revision as of 09:15, 9 November 2024
Contents
[hide]Problem
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
Given the information above, we start with the 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 remaining, 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
Solution 4: Answer choices
We can rewrite the question as an algebraic equation: , where represents the total amount of points and the amount of points the other players scored. From there, we add the two fractions to get . Subtracting from both sides, we get . We multiply each side by to get rid of the denominator, in which we get . Now let’s think of this logically. This equation is telling us that if you add and times the amount of points scored by the extra players, you get times the amount of points total. And since we have to have a whole number of points total, this means that must be divisible by . Plugging in all the answer choices for , we find that the only answer that makes divisible by is .
~ilee0820
Video Solution by Math-X (Learn to do this under a minute!!!)
https://youtu.be/IgpayYB48C4?si=JjQHbrlBpeox9TFq&t=7063
~Math-X
https://www.youtube.com/watch?v=fKjmw_zzCUU
- Happytwin
https://www.youtube.com/watch?v=jE-7Se7ay1c
Associated video
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
~savannahsolver
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
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.