Question

Raiders 和 Wildcats 的一场高中篮球赛在第一节结束时打平。Raiders 四节每节得分构成一个递增等比数列，Wildcats 四节每节得分构成一个递增等差数列。第四节结束时 Raiders 以一分获胜。两队得分都不超过 $100$ 。两队上半场总共得了多少分？

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?

$30$

$31$

$32$

$33$

$34$

答案：E

知识点：等比数列等差数列丢番图方程

难度评级：2180

解答：

设两队第一节得分均为 $a$ 。设 Raiders 的公比为 $r$ ，Wildcats 的公差为 $d$ 。

将 $r=m/n$ 写成最简分数。由于 Raiders 四节得分都是整数， $a=n^3A$ ，其中 $A$ 为正整数，且总分为 $A(n^3+n^2m+nm^2+m^3)\le100$ ，所以只需检查 $\dfrac{3}{2},2,3,4$ 。

对 $r=\dfrac{3}{2}$ 或 $r=4$ ，Raiders 的可行总分无法使 Wildcats 总分成为 $4a+6d$ 的形式。对 $r=3$ ，方程 $40a=4a+6d+1$ 模 $6$ 不可能。

对 $r=2$ ，方程为 $15a=4a+6d+1$ ，所以 $11a=6d+1$ 。总分不超过 $100$ 的唯一正整数解是 $a=5,d=9$ 。

上半场总分为 $5+10+5+14=34$ 。

所以正确答案是 E。

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 $r=m/n$ 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.

2010 AMC 10B 第 24 题

答案：E

难度评级：2180

解答：

其他年份的第 24 题