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

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==Solution 1==
 
==Solution 1==
 
Since <math>\frac{\text{total points}}{4}</math> and <math>\frac{2(\text{total points})}{7}</math> are integers, we have <math>28 | \text{total points}</math>. We see that the number of points scored by the other team members is less than or equal to <math>14</math> and greater than or equal to <math>0</math>. We let the total number of points be <math>t</math> and the total number of points scored by the other team members, which means that <math>\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14</math>, which means <math>15 \le \frac{13t}{28} \le 29</math>. The only value of <math>t</math> that satisfies all conditions listed is <math>56</math>, so <math>x = \boxed{11}</math>. - juliankuang
 
Since <math>\frac{\text{total points}}{4}</math> and <math>\frac{2(\text{total points})}{7}</math> are integers, we have <math>28 | \text{total points}</math>. We see that the number of points scored by the other team members is less than or equal to <math>14</math> and greater than or equal to <math>0</math>. We let the total number of points be <math>t</math> and the total number of points scored by the other team members, which means that <math>\frac{t}{4} + \frac{2t}{7} + 15 + x = t \quad \implies \quad 0 \le \frac{13t}{28} - 15 = x \le 14</math>, which means <math>15 \le \frac{13t}{28} \le 29</math>. The only value of <math>t</math> that satisfies all conditions listed is <math>56</math>, so <math>x = \boxed{11}</math>. - juliankuang
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==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>.
  
 
==See Also==
 
==See Also==

Revision as of 04:20, 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 $28x+420=13t$. Since $t$ is necessarily divisible by 7, let $t=7u$ and divide by 7 to obtain $4x + 60 = 13u$. Then $4\mid u$, and we see $u=4$ is invalid as this implies $x < 0$, but $u=8$ works, giving $x=\boxed{\textbf{(B)} 11}$.

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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