AMC10 2010 B

AMC10 2010 B · Q24

AMC10 2010 B · Q24. It mainly tests Sequences & recursion (algebra), Arithmetic misc.

A high school basketball game between the Raiders and the 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?

Raiders队与Wildcats队的高中篮球比赛第一节结束时打平。Raiders队在四个节得分构成递增等比数列，Wildcats队构成递增等差数列。第四节结束时，Raiders队以1分优势获胜。两队均未得分超过100分。两队上半场总得分是多少？

(A) 30 30

(B) 31 31

(D) 33 33

(E) 34 34

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): The Raiders’ score was $a(1+r+r^2+r^3)$, where $a$ is a positive integer and $r>1$. Because $ar$ is also an integer, $r=m/n$ for relatively prime positive integers $m$ and $n$ with $m>n$. Moreover $ar^3=a\cdot \frac{m^3}{n^3}$ is an integer, so $n^3$ divides $a$. Let $a=n^3A$. Then the Raiders’ score was $R=A(n^3+mn^2+m^2n+m^3)$, and the Wildcats’ score was $R-1=a+(a+d)+(a+2d)+(a+3d)=4a+6d$ for some positive integer $d$. Because $A\ge 1$, the condition $R\le 100$ implies that $n\le 2$ and $m\le 4$. The only possibilities are $(m,n)=(4,1),(3,2),(3,1)$, or $(2,1)$. The corresponding values of $R$ are, respectively, $85A,65A,40A$, and $15A$. In the first two cases $A=1$, and the corresponding values of $R-1$ are, respectively, $64=32+6d$ and $84=4+6d$. In neither case is $d$ an integer. In the third case $40A=40a=4a+6d+1$ which is impossible in integers. In the last case $15a=4a+6d+1$, from which $11a=6d+1$. The only solution in positive integers for which $4a+6d\le 100$ is $(a,d)=(5,9)$. Thus $R=5+10+20+40=75$, $R-1=5+14+23+32=74$, and the number of points scored in the first half was $5+10+5+14=34$.

答案（E）：突袭者队（Raiders）的得分为 $a(1+r+r^2+r^3)$，其中 $a$ 是正整数且 $r>1$。因为 $ar$ 也是整数，所以 $r=m/n$，其中 $m,n$ 为互素的正整数且 $m>n$。另外 $ar^3=a\cdot \frac{m^3}{n^3}$ 是整数，因此 $n^3$ 整除 $a$。令 $a=n^3A$。则突袭者队得分为 $R=A(n^3+mn^2+m^2n+m^3)$，而野猫队（Wildcats）的得分为 $R-1=a+(a+d)+(a+2d)+(a+3d)=4a+6d$，其中 $d$ 为某个正整数。由于 $A\ge 1$，条件 $R\le 100$ 推出 $n\le 2$ 且 $m\le 4$。唯一可能为 $(m,n)=(4,1),(3,2),(3,1)$ 或 $(2,1)$。对应的 $R$ 值分别为 $85A,65A,40A,15A$。前两种情形中 $A=1$，相应的 $R-1$ 分别为 $64=32+6d$ 与 $84=4+6d$，但两种情况下 $d$ 都不是整数。第三种情形中 $40A=40a=4a+6d+1$，在整数中不可能成立。最后一种情形中 $15a=4a+6d+1$，从而 $11a=6d+1$。在满足 $4a+6d\le 100$ 的正整数解中，唯一解为 $(a,d)=(5,9)$。因此 $R=5+10+20+40=75$，$R-1=5+14+23+32=74$，上半场得分为 $5+10+5+14=34$。

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