Question

去年 Isabella 参加了 $7$ 次数学考试，得到了 $7$ 个不同的分数，每个分数都是 $91$ 到 $100$ 之间（含端点）的整数。每次考试后，她都注意到到目前为止考试分数的平均数是整数。她第七次考试得了 $95$ 分。她第六次考试得了多少分？

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100,$ inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95.$ What was her score on the sixth test?

$92$

$94$

$96$

$98$

$100$

答案：E

知识点：整除性平均数极限情形界定

难度评级：2300

解答：

设七次总分为 $S$ 。因为七次后的平均数是整数， $S$ 可被 $7$ 整除。七个不同分数都在 $91$ 到 $100$ 之间，所以 $91+92+\cdots+97$ $\le S\le$ $94+95+\cdots+100$ 。

因此 $658\le S\le 679$ ，可能的 $7$ 的倍数为 $658,665,672,679$ 。由于第七次分数是 $95$ ，前六次总分为 $S-95$ ，它必须能被 $6$ 整除。这迫使 $S=665$ 。

因此前六次总分为 $570$ 。前五次平均数也是整数，所以前五次总分能被 $5$ 整除。因此第六次分数也必须能被 $5$ 整除。由于第七次已经是 $95$ ，且所有分数不同，第六次分数只能是 $100$ 。所以正确答案是 E。

Let $S$ be the sum of all seven scores. Since all seven averages were integers, $S$ is divisible by $7$ . Also the seven distinct scores are between $91$ and $100$ , so $91+92+\cdots+97$ $\le S\le$ $94+95+\cdots+100$ .

Thus $658\le S\le 679$ , and the possible multiples of $7$ are $658,665,672,679$ . Since the seventh score is $95$ , the first six scores sum to $S-95$ , which must be divisible by $6$ . This forces $S=665$ .

The first six scores sum to $570$ . The first five-score average was also an integer, so the sum of the first five scores is divisible by $5$ . Therefore the sixth score is divisible by $5$ . Since the seventh score is already $95$ and all scores are distinct, the sixth score is $100$ . Thus, E is the correct answer.

2017 AMC 10B 第 25 题

答案：E

难度评级：2300

解答：

其他年份的第 25 题