AMC8 2008

AMC8 2008 · Q15

AMC8 2008 · Q15. It mainly tests Averages (mean), Remainders & modular arithmetic.

In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

在特蕾莎的前8场篮球比赛中，她得分分别是7、4、3、6、8、3、1和5分。在第九场比赛中，她得分少于10分，且九场比赛的场均得分是整数。同样，在第十场比赛中，她得分少于10分，且十场比赛的场均得分也是整数。第九场和第十场比赛得分的乘积是多少？

(A) 35 35

(B) 40 40

(D) 56 56

(E) 72 72

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The sum of the points Theresa scored in the first $8$ games is $37$. After the ninth game, her point total must be a multiple of $9$ between $37$ and $37+9=46$, inclusive. The only such point total is $45=37+8$, so in the ninth game she scored $8$ points. Similarly, the next point total must be a multiple of $10$ between $45$ and $45+9=54$. The only such point total is $50=45+5$, so in the tenth game she scored $5$ points. The product of the number of points scored in Theresa’s ninth and tenth games is $8\cdot 5=40$.

答案（B）：Theresa 在前 $8$ 场比赛中得到的总分是 $37$。打完第九场后，她的总分必须是介于 $37$ 和 $37+9=46$（含）之间的 $9$ 的倍数。唯一满足的总分是 $45=37+8$，因此第九场她得了 $8$ 分。同理，下一次总分必须是介于 $45$ 和 $45+9=54$ 之间的 $10$ 的倍数。唯一满足的总分是 $50=45+5$，因此第十场她得了 $5$ 分。Theresa 第九场和第十场得分的乘积为 $8\cdot 5=40$。

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