# Difference between revisions of "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 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 $(5)+(10)+(5)+(14)=\boxed{\textbf{(E)}\ 34}$.

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