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Difference between revisions of "2010 AMC 10B Problems/Problem 24"
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*<math>        n=3, a=[1,2]</math>
 
*<math>        n=3, a=[1,2]</math>
 
*<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 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 <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 a=6 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 <math>(a)+(an)+(a)+(a+m)=(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 23:23, 10 January 2015

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

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

We have $a+an+an^2+an^3=4a+6m+1$ Manipulating this, 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=6 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 $(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}$

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

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