Difference between revisions of "2010 AMC 12B Problems/Problem 19"
(+{{solution}}) |
CRICKET229 (talk | contribs) (→Solution) |
||
Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
+ | |||
+ | Let <math>a,ar,ar^{2},ar^{3}</math> be the quarterly scores for the Raiders. We know that The Raiders and Wildcats both scored the same number of points in the first quarter so let <math>a,a+d,a+2d,a+3d</math> be quarterly scores for the Raiders. The sum of the Raiders scores is <math>a(1+r+r^{2}+r^{3})</math> and the sum of the Wildcats scores is <math>4a+6d</math>. Now we can narrow our search for the values of <math>a,d</math>, and <math>r</math>. Because points are always measured in positive integers, we can conclude that <math>a</math> and <math>d</math> are positive integers. We can also conclude that <math>r</math> is a positive integer by writing down the equation: | ||
+ | |||
+ | <cmath>a(1+r+r^{2}+r^{3})=4a+6d+1</cmath> | ||
+ | |||
+ | Because both the left hand and right hand sides must be integers, we have that <math>r+r^{2}+r^{3}</math> is a positive integer and due to our restrictions, it follows that <math>r</math> must be a positive integer as well. Now we can start trying out some values of <math>r</math>. We start at <math>r=1</math> which gives that <math>6d=-1</math> which is a contradiction. Next we try <math>r=2</math>, which gives | ||
+ | |||
+ | <cmath>15a=4a+6d+1</cmath> | ||
+ | <cmath>11a=6d+1</cmath> | ||
+ | |||
+ | We need the smallest multiple of <math>11</math> (to satisfy the <100 condition) that is <math>\equiv 1 \pmod{6}</math>. We see that this is <math>\boxed{55}</math>, and therefore <math>a=5</math> and <math>d=9</math>. | ||
+ | |||
+ | So the Raiders first two scores were <math>5</math> and <math>10</math> and the Wildcats first two scores were <math>5</math> and <math>14</math>. | ||
+ | |||
+ | <cmath>5+10+5+14=34 \longrightarrow $\boxed{\textbf{(E)}}</cmath> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}} | {{AMC12 box|year=2010|num-b=18|num-a=20|ab=B}} |
Revision as of 17:00, 31 May 2011
This problem needs a solution. If you have a solution for it, please help us out by adding it.
Problem 19
A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?
Solution
Let be the quarterly scores for the Raiders. We know that The Raiders and Wildcats both scored the same number of points in the first quarter so let
be quarterly scores for the Raiders. The sum of the Raiders scores is
and the sum of the Wildcats scores is
. Now we can narrow our search for the values of
, and
. Because points are always measured in positive integers, we can conclude that
and
are positive integers. We can also conclude that
is a positive integer by writing down the equation:
Because both the left hand and right hand sides must be integers, we have that is a positive integer and due to our restrictions, it follows that
must be a positive integer as well. Now we can start trying out some values of
. We start at
which gives that
which is a contradiction. Next we try
, which gives
We need the smallest multiple of (to satisfy the <100 condition) that is
. We see that this is
, and therefore
and
.
So the Raiders first two scores were and
and the Wildcats first two scores were
and
.
\[5+10+5+14=34 \longrightarrow $\boxed{\textbf{(E)}}\] (Error compiling LaTeX. Unknown error_msg)
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 AMC 12 Problems and Solutions |