1982 AMC 12 第 28 题

先试着解答 1982 AMC 12 第 28 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 1982 AMC 12 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

28.

黑板上写有一组从 11 开始的连续正整数。擦去其中一个数后,其余各数的平均数(算术平均值)为 3571735\frac7{17}。擦去的是哪个数?
A set of consecutive positive integers beginning with 11 is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is 35717.35\frac7{17}. What number was erased?
66
77
88
99
无法确定
can not be determined
答案:B
知识点:平均数等差数列整除性
小提示:
若写出的最大数为 nn,则剩余数的个数为 n−1n-1
If the last written number is n,n, the remaining count is n−1n-1
大提示:
利用被擦去的数满足 1≤e≤n1\le e\le n,缩小 nn 的可能取值范围
Use the bounds 1≤e≤n1\le e\le n on the erased value to narrow the possible values of nn
解答:
新的平均数 60217\frac{602}{17} 小于原平均数 n+12\frac{n+1}{2},但两者之差小于 nn−1\frac{n}{n-1},因而 n=69n=69 或 7070。又因为剩余各数之和 (n−1)(60217)(n-1)(\frac{602}{17}) 必须为整数,所以 n−1n-1 是 1717 的倍数,从而 n=69n=69。擦去的数为 2415−68⋅60217=2415−2408=7。 \begin{aligned} 2415-68\cdot\frac{602}{17} &=2415-2408\\ &=7 \end{aligned}\text{。}
因此,正确答案为 B。
The new average 60217\frac{602}{17} is below the original average n+12\frac{n+1}{2} but differs from it by less than nn−1,\frac{n}{n-1}, forcing n=69n=69 or 70.70. Since the remaining sum (n−1)(60217)(n-1)(\frac{602}{17}) must be integral, n−1n-1 is divisible by 17,17, so n=69.n=69. The erased number is 2415−68⋅60217=2415−2408=7. \begin{aligned} 2415-68\cdot\frac{602}{17} &=2415-2408\\ &=7. \end{aligned}
Therefore, the correct answer is B.
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