2017 AMC 12B 第 21 题

Question

去年 Isabella 参加了

7

次数学测验，得到

7

个互不相同的分数，每个分数都是

91

到

100

之间（含端点）的整数。每次测验后，她都注意到到目前为止所有测验分数的平均数是整数。她第七次测验的分数是

95

。她第六次测验的分数是多少？

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：2040

小提示：

所有

7

个分数之和

S

是

7

的倍数，并且介于

658

和

679

之间

The sum

S

of all

7

scores is a multiple of

7

and lies between

658

and

679

大提示：

还要有

S - 95

是

6

的倍数；然后利用前

5

个分数之和是

5

的倍数，限制第六个分数的余数。

Also

S - 95

is a multiple of

6;

then the first

5

scores sum to a multiple of

5,

forcing the sixth score’s residue

解答：

令

S

为七个分数之和。则

S

是

7

的倍数且

658 \le S \le 679

，所以

S \in \{658, 665, 672, 679\}

。因为前六次测验后的平均数是整数，

S - 95

必须是

6

的倍数，这迫使

S = 665

。因此前六个分数之和为

570

，它是

5

的倍数。前五次测验后的平均数也是整数，所以前五个分数之和是

5

的倍数，于是第六个分数也是

5

的倍数。由于所有分数互不相同且第七个分数为

95

，第六个分数必须是

100

。

所以正确答案是 E。

Let

S

be the sum of all seven scores. Then

S

is a multiple of

7

with

658 \le S \le 679,

so

S \in \{658, 665, 672, 679\}.

Since the average after six tests is an integer,

S - 95

is a multiple of

6,

which forces

S = 665.

Then the first six scores sum to

570,

a multiple of

5;

the average after five tests is an integer, so the first five scores also sum to a multiple of

5,

making the sixth score a multiple of

5.

Since all scores differ and the seventh is

95,

the sixth must be

100.

Thus, the correct answer is E.

2017 AMC 12B 第 21 题

答案：E

难度评级：2040

小提示：

大提示：

解答：

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