Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2010 AMC 10B Problems/Problem 24"

m (→‎Solution)
(→‎Solution)
Line 4: Line 4:
 
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math>
 
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34</math>
  
−
== Solution ==
+
== Solution 1 ==
 
Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math>
 
Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math>
  
Line 15: Line 15:
 
*<math>          n=4, a=1</math>
 
*<math>          n=4, a=1</math>
 
Checking each of these cases individually back into the equation <math>a+an+an^2+an^3=4a+6m+1</math>, we see that only when <math>a=5</math> and <math>n=2</math>, we get an integer value for <math>m</math>, which is <math>9</math>. The original question asks for the first half scores summed, so we must find <math>(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}</math>
 
Checking each of these cases individually back into the equation <math>a+an+an^2+an^3=4a+6m+1</math>, we see that only when <math>a=5</math> and <math>n=2</math>, we get an integer value for <math>m</math>, which is <math>9</math>. The original question asks for the first half scores summed, so we must find <math>(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}</math>
 +
 +
== Solution 2 ==
 +
 +
As above, represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math>.
 +
 +
Note that the Wildcat's sum, <math>4a+6m</math>, is even. Therefore, since the Wildcat's sum is one less than the Raiders, the Raider's team's score should be odd. But if all of <math>(a, an, an^2, an^3)</math> are of the same parity, the sum will be even. If <math>a</math> is even, then the rest of the scores will be even, so clearly <math>a</math> is odd. Then, <math>n</math> is even.
 +
 +
But if <math>n=4</math>, only <math>a=1</math> satisfies the requirement that the total score of each team is less than <math>100</math>. We can test this out and see it doesn't work.
 +
 +
Therefore, <math>n=2</math>. If we try <math>a=(1,3,5)</math>, we quickly see only <math>a=5</math> satisfies all of the conditions. Therefore, our team's scores are <math>(5,10,20,40)</math> and <math>(5,14,23,32)</math>, and the answer is <math>(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}
 
{{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:02, 23 January 2018

Problem

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 $100$ points. What was the total number of points scored by the two teams in the first half?

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution 1

Represent the teams' scores as: $(a, an, an^2, an^3)$ and $(a, a+m, a+2m, a+3m)$

We have $a+an+an^2+an^3=4a+6m+1$ Factoring out the $a$ from the left side of the equation, we can get $a(1+n+n^2+n^3)=4a+6m+1$, or $a(n^4-1)/(n-1)=4a+6m+1$

Since both are increasing sequences, $n>1$. We can check cases up to $n=4$ because when $n=5$, we get $156a>100$. When

  • $n=2, a=[1,6]$
  • $n=3, a=[1,2]$
  • $n=4, a=1$

Checking each of these cases individually back into the equation $a+an+an^2+an^3=4a+6m+1$, we see that only when $a=5$ and $n=2$, we get an integer value for $m$, which is $9$. The original question asks for the first half scores summed, so we must find $(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}$

Solution 2

As above, represent the teams' scores as: $(a, an, an^2, an^3)$ and $(a, a+m, a+2m, a+3m)$.

Note that the Wildcat's sum, $4a+6m$, is even. Therefore, since the Wildcat's sum is one less than the Raiders, the Raider's team's score should be odd. But if all of $(a, an, an^2, an^3)$ are of the same parity, the sum will be even. If $a$ is even, then the rest of the scores will be even, so clearly $a$ is odd. Then, $n$ is even.

But if $n=4$, only $a=1$ satisfies the requirement that the total score of each team is less than $100$. We can test this out and see it doesn't work.

Therefore, $n=2$. If we try $a=(1,3,5)$, we quickly see only $a=5$ satisfies all of the conditions. Therefore, our team's scores are $(5,10,20,40)$ and $(5,14,23,32)$, and the answer is $(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}$.

See also

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_AMC_10B_Problems/Problem_24&oldid=89949"