Difference between revisions of "2019 AMC 8 Problems/Problem 23"
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==Solution 2== | ==Solution 2== | ||
− | 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+ | + | 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 03:24, 22 November 2019
Contents
[hide]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, which means that
, which means
. The only value of
that satisfies all conditions listed is
, so
. - juliankuang
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
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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