2010 AMC 10B Problem 24

Let both teams score $a$ in the first quarter. Raiders: $a, ar, ar^2, ar^3$ with ratio $r > 1$; Wildcats: $a, a+d, a+2d, a+3d$ with $d > 0$. The totals satisfy
$$a(1 + r + r^2 + r^3) = 4a + 6d + 1 \le 100.$$

Since all Raiders' scores are integers, if $r = \frac{p}{q}$ in lowest terms then $q^3 \mid a$. Check the possible ratios under the $100$-point cap:

• $r = 2$: total $15a \le 100$, so $a \le 6$.
• $r = 3$: total $40a$, so $a \le 2$.
• $r = 4$: total $85a$, so $a = 1$.
• $r = \frac32$: $a = 8k$, total $65k$, so $a = 8$, total $65$.
• Any other ratio ($5$, $\frac43$, $\frac52$, ...) exceeds $100$.

Now require $6d = (\text{Raiders total}) - 1 - 4a$ to be a positive multiple of $6$:
- $r = 2$: $11a - 1$ must be divisible by $6$; for $a = 1, \dots, 6$ the values are $10, 21, 32, 43, 54, 65$, and only $a = 5$ works, with $d = 9$.
- $r = 3$: $36a - 1$ is never a multiple of $6$.
- $r = 4$, $a = 1$: $80$ is not a multiple of $6$.
- $r = \frac32$, $a = 8$: $65 - 1 - 32 = 32$, not a multiple of $6$.

So the Raiders scored $5, 10, 20, 40$ (total $75$) and the Wildcats $5, 14, 23, 32$ (total $74$). First-half points: $5 + 10 + 5 + 14 = 34$.

The answer is $\boxed{\textbf{(E)}\ 34}$.

A cap on the total makes a geometric sequence very rigid: $1 + r + r^2 + r^3$ grows fast, so only a handful of $(a, r)$ pairs survive, and the integer-terms condition rules out most fractional ratios. Once the geometric side is enumerated, the arithmetic side is a divisibility check on $4a + 6d$.

• Trying ratio 3/2 with a first term not divisible by 8, or forgetting that the two teams share the same first-quarter score a.
• Answering with the Raiders' first half ($15$) or the total game score, or checking only $r = 2$ without justifying that other ratios fail.