2010 AMC 10B Problems/Problem 24

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

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2010 AMC 10B Problems/Problem 24

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