2010 AMC 12B Problem 19

Let the Raiders score $a, ar, ar^2, ar^3$ (increasing geometric, $r\gt1$) and the Wildcats $a, a+d, a+2d, a+3d$ (increasing arithmetic), tied in the first quarter at $a.$ Write $r=\frac{m}{n}$ in lowest terms, with $m\gt n.$ Since $ar^3$ is an integer, $a$ is divisible by $n^3;$ put $a=A n^3.$ The Raiders’ total is $$R=A(n^3+mn^2+m^2n+m^3).$$ Since $m^3\lt R\le100,$ we have $m\le4.$ The pair $(4,3)$ already makes the parenthesized sum $175,$ so the only possible coprime pairs $(m,n)$ are $(2,1),(3,1),(3,2),(4,1).$ The corresponding base values of $\frac{R}{A}$ are $15,40,65,85.$ For $(4,1)$ and $(3,2),$ the bound forces $A=1,$ but $R-1=4a+6d$ gives a nonintegral $d.$ For $(3,1),$ it would give $36A-1=6d,$ which is impossible modulo $6.$ For $(m,n)=(2,1),$ we have $R=15A$ and $a=A,$ so $11A-1=6d.$ Thus $A\equiv5\pmod6,$ and $R\le100$ forces $A=5.$ Then $d=9,$ giving Raiders scores $5,10,20,40$ and Wildcats scores $5,14,23,32.$ The Raiders win $75$ to $74.$ The first-half total is $(5+10)+(5+14)=34.$ Thus, the correct answer is E.