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

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==Solution 2==
 
==Solution 2==
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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+420=13t</math>. Since <math>t</math> is necessarily divisible by 7, let <math>t=7u</math> and divide by 7 to obtain <math>4x + 60 = 13u</math>. Then <math>4\mid u</math>, and we see <math>u=4</math> is invalid as this implies <math>x < 0</math>, but <math>u=8</math> works, giving <math>x=\boxed{\textbf{(B)} 11}</math>. -scrabbler94
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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
  
 
==See Also==
 
==See Also==

Revision as of 04:24, 22 November 2019

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

Since $\frac{\text{total points}}{4}$ and $\frac{2(\text{total points})}{7}$ are integers, we have $28 | \text{total points}$. We see that the number of points scored by the other team members is less than or equal to $14$ and greater than or equal to $0$. We let the total number of points be $t$ and the total number of points scored by the other team members, which means that $\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14$, which means $15 \le \frac{13t}{28} \le 29$. The only value of $t$ that satisfies all conditions listed is $56$, so $x = \boxed{11}$. - juliankuang

Solution 2

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

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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