AMC12 2017 B

AMC12 2017 B · Q21

AMC12 2017 B · Q21. It mainly tests Averages (mean), GCD & LCM.

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?

去年Isabella参加了7次数学测验，得到了7个不同的分数，每个分数都是91到100之间的整数。每次测验后，她注意到她的测验平均分都是整数。她的第七次测验分数是95。她的第六次测验分数是多少？

(A) 92 92

(B) 94 94

(D) 98 98

(E) 100 100

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $S$ be the sum of Isabella’s 7 scores. Then $S$ is a multiple of 7, and $658 = 91 + 92 + 93 + \cdots + 97 \le S \le 94 + 95 + 96 + \cdots + 100 = 679,$ so $S$ is one of 658, 665, 672, or 679. Because $S - 95$ is a multiple of 6, it follows that $S = 665$. Thus the sum of Isabella’s first 6 scores was $665 - 95 = 570$, which is a multiple of 5, and the sum of her first 5 scores was also a multiple of 5. Therefore her sixth score must have been a multiple of 5. Because her seventh score was 95 and her scores were all different, her sixth score was 100. One possible sequence of scores is 91, 93, 92, 96, 98, 100, 95.

答案（E）：设 $S$ 为伊莎贝拉 7 次成绩之和。则 $S$ 是 7 的倍数，并且 $658 = 91 + 92 + 93 + \cdots + 97 \le S \le 94 + 95 + 96 + \cdots + 100 = 679,$ 所以 $S$ 只能是 658、665、672 或 679 之一。由于 $S - 95$ 是 6 的倍数，可得 $S = 665$。因此伊莎贝拉前 6 次成绩之和为 $665 - 95 = 570$，它是 5 的倍数；而她前 5 次成绩之和也同样是 5 的倍数。所以她第六次成绩必须是 5 的倍数。又因为她第七次成绩是 95 且各次成绩互不相同，所以她第六次成绩为 100。一种可能的成绩序列是 91，93，92，96，98，100，95。

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