2000 AMC 12 第 9 题

Question

Walter 老师给一个五人数学班考试。她把成绩按随机顺序输入电子表格，表格在每输入一个成绩后重新计算班级平均分。Walter 老师注意到每次输入后，平均分总是整数。五个成绩按升序为

71

，

76

，

80

，

82

和

91

。Walter 老师最后输入的成绩是多少？

Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were

71,

76,

80,

82,

and

91.

What was the last score Mrs. Walter entered?

71

76

80

82

91

答案：C

知识点：整除性模运算平均数

难度评级：1580

小提示：

前

k

个输入成绩的和必须能被

k

整除。

The sum of the first

k

scores entered must be divisible by

k

大提示：

先用能被

3

整除，再用能被

4

整除；这些成绩模

3

的余数是

2, 1, 2, 1, 1

。

Use divisibility by

3

then by

4

; the residues of the scores modulo

3

are

2, 1, 2, 1, 1

解答：

总和为

71 + 76 + 80 + 82 + 91 = 400

，可被

5

整除。前三个成绩的和必须能被

3

整除。

模

3

，这些成绩的余数是

2, 1, 2, 1, 1

。唯一和为

3

的倍数的三元组是

76 + 82 + 91 = 249

，所以它们是前三个成绩（其中

91

第三个，因为前两个

76

和

82

，必须奇偶性相同）。

因为

249 \equiv 1 \pmod 4

，第四个成绩必须

\equiv 3 \pmod 4

，即

71

。于是最后输入的是

80

。

因此，正确答案是 C。

The total is

71 + 76 + 80 + 82 + 91 = 400,

which is divisible by

5.

The sum of the first three scores must be divisible by

3.

Modulo

3,

the scores are

2, 1, 2, 1, 1.

The only triple summing to a multiple of

3

is

76 + 82 + 91 = 249,

so these are the first three (with

91

third, since the first two,

76

and

82,

must have equal parity).

Since

249 \equiv 1 \pmod 4,

the fourth score must be

\equiv 3 \pmod 4,

which is

71.

That leaves

80

as the last score entered.

Thus, the correct answer is C.

2000 AMC 12 第 9 题

答案：C

难度评级：1580

小提示：

大提示：

解答：

其他年份的第 9 题