Question

Theresa 前 $8$ 场篮球比赛分别得了 $7, 4, 3, 6, 8, 3, 1$ 和 $5$ 分。第九场她得分少于 $10$ 分，并且九场比赛的场均得分是整数。类似地，第十场她得分少于 $10$ 分，并且 $10$ 场比赛的场均得分也是整数。她第九场和第十场得分的乘积是多少？

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?

$\ 35$

$\ 40$

$\ 48$

$\ 56$

$\ 72$

答案：B

知识点：平均数整除性

难度评级：1310

解答：

她前 $8$ 场的得分和为 $37$ 。因为前 $9$ 场的平均分是整数，所以前 $9$ 场的总分是 $9$ 的倍数。

由于第九场得分少于 $10$ ，前 $9$ 场的总分在 $37$ 到 $47$ 之间，并且是 $9$ 的倍数，因此总分为 $45$ 。所以第 $9$ 场得分为 $45-37=8$ 。

Theresa 前 $9$ 场的总分是 $45$ 。因为前 $10$ 场的平均分是整数，所以前 $10$ 场的总分是 $10$ 的倍数。

由于第十场得分少于 $10$ ，前 $10$ 场的总分在 $45$ 到 $55$ 之间，并且是 $10$ 的倍数，因此总分为 $50$ 。所以第 $10$ 场得分为 $50-45=5$ 。

因此乘积为 $8\cdot 5=40$ 。

所以正确答案是 B。

The sum of her first $8$ scores is $37.$ Since the average of the first $9$ scores is an integer, the sum of the first $9$ scores is a multiple of $9.$

Since the score is less than $10,$ the sum of the scores after $9$ games is between $37$ and $47,$ and is a multiple of $9,$ making the sum $45.$ Thus, the score of the $9$ th game is $45-37=8.$

The sum of Theresa's first $9$ scores is $45.$ Since the average of the first $10$ scores is an integer, the sum of the first $10$ scores is a multiple of $10.$

Since the score is less than $10,$ the sum of the scores after $10$ games is between $45$ and $55,$ and is a multiple of $10,$ making the sum $50.$ Thus, the score of the $10$ th game is $50-45=5.$

Therefore, their product is $8\cdot 5=40.$

Thus, the answer is B .

2008 AMC 8 第 15 题

答案：B

难度评级：1310

解答：

其他年份的第 15 题