2010 AMC 10B Problem 24

Let the first-quarter score for each team be $a.$ Let the Raiders have common ratio $r$ and let the Wildcats have common difference $d.$ Write $r=\frac{m}{n}$ in lowest terms. Since the Raiders’ four quarter scores are integers, $a=n^3A$ for some positive integer $A$, and their total is $A(n^3+n^2m+nm^2+m^3)\le100$. Thus the only possible ratios are $\dfrac{3}{2},2,3,4$. For $r=\dfrac{3}{2}$ or $r=4$, the only possible Raiders totals give Wildcats totals that are not of the form $4a+6d$. For $r=3$, the equation $40a=4a+6d+1$ is impossible modulo $6$. For $r=2$, the equation is $15a=4a+6d+1$, so $11a=6d+1$. The only positive solution with total at most $100$ is $a=5,d=9$. The first-half total is $5+10+5+14=34.$ Thus, E is the correct answer.