2010 AMC 10B Problem 24

Both teams scored the same amount, call it $a$, in the first quarter.

• Wildcats: $a,\ a + d,\ a + 2d,\ a + 3d$, with $d$ a positive whole number. Total: $4a + 6d$.
• Raiders: $a,\ ar,\ ar^2,\ ar^3$, with ratio $r$ bigger than $1$. Their total is $1$ more than the Wildcats' total.

So the Raiders' total minus $1$ minus $4a$ must equal $6d$: a positive multiple of $6$. The Raiders' side has far fewer options, so list those first.

Step 1: list the possible Raiders' scores (total at most $100$).

1. Ratio $2$: $a, 2a, 4a, 8a$, total $15a$. So $a$ is at most $6$.
2. Ratio $3$: total $a + 3a + 9a + 27a = 40a$. So $a$ is $1$ or $2$.
3. Ratio $4$: total $85a$, so $a = 1$.
4. Ratio $5$ or more: even with $a = 1$ the total is at least $1 + 5 + 25 + 125 = 156$. Too big.
5. A fraction ratio such as $\frac32$: every quarter score must be whole, and the fourth score is $a \cdot \frac{27}{8}$, so $a$ must be a multiple of $8$. With $a = 8$ the scores are $8, 12, 18, 27$, total $65$; $a = 16$ doubles that to $130$, too big.
6. Other fractions fail too. With a bottom of $2$, the next choice is $\frac52$, and $a = 8$ gives $8, 20, 50, 125$, too big. With a bottom of $3$ or more (like $\frac43$), the fourth score has a $27$ or more on the bottom, so $a$ is at least $27$; then all four scores are at least $27$ and the total is over $100$.

Step 2: test each option. We need (Raiders' total) $- 1 - 4a$ to be a positive multiple of $6$.

Step 3: read off the game. Only $a = 5$, ratio $2$, $d = 9$ works.

• Raiders: $5, 10, 20, 40$, total $75$.
• Wildcats: $5, 14, 23, 32$, total $74$. Raiders win by $1$.

The first half is the first two quarters: $5 + 10 + 5 + 14 = 34$.

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

A growing geometric sequence gets big very fast, so a limit of $100$ points leaves only a handful of possibilities, and the whole-number scores rule out most fraction ratios. Once the short list is written down, the arithmetic side is just a check of whether a number is a multiple of $6$. When one side of a problem has few options, list that side first.

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 $\boxed{\textbf{(E)}\ 34}$.