AMC10 2000 A

AMC10 2000 A · Q14

AMC10 2000 A · Q14. It mainly tests Remainders & modular arithmetic.

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?

沃尔特夫人给一个五名学生的数学班考试。她随机顺序将分数输入电子表格，每次输入一个分数后电子表格重新计算班级平均分。沃尔特夫人注意到，每次输入分数后，平均分总是整数。分数（按升序排列）是71、76、80、82和91。沃尔特夫人最后输入的分数是什么？

(A) 71 71

(B) 76 76

(D) 82 82

(E) 91 91

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Note that the integer average condition means that the sum of the scores of the first $n$ students is a multiple of $n$. The scores of the first two students must be both even or both odd, and the sum of the scores of the first three students must be divisible by $3$. The remainders when $71$, $76$, $80$, $82$, and $91$ are divided by $3$ are $2$, $1$, $2$, $1$, and $1$, respectively. Thus the only sum of three scores divisible by $3$ is $76+82+91=249$, so the first two scores entered are $76$ and $82$ (in some order), and the third score is $91$. Since $249$ is $1$ larger than a multiple of $4$, the fourth score must be $3$ larger than a multiple of $4$, and the only possible is $71$, leaving $80$ as the score of the fifth student.

答案（C）：注意，整数平均数条件意味着前 $n$ 个学生的分数之和是 $n$ 的倍数。前两名学生的分数必须同为偶数或同为奇数，并且前三名学生分数之和必须能被 $3$ 整除。$71$、$76$、$80$、$82$ 和 $91$ 分别除以 $3$ 的余数是 $2$、$1$、$2$、$1$ 和 $1$。因此，三个分数之和能被 $3$ 整除的唯一组合是 $76+82+91=249$，所以前两次输入的分数是 $76$ 和 $82$（顺序不定），第三个分数是 $91$。由于 $249$ 比某个 $4$ 的倍数大 $1$，第四个分数必须比某个 $4$ 的倍数大 $3$，唯一可能是 $71$，于是第五名学生的分数为 $80$。

Topics

NT.004 Remainders & modular arithmetic