2016 AMC 12B 第 7 题

先试着解答 2016 AMC 12B 第 7 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 2016 AMC 12B 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。

7.

Josh 写下数字 11,22,33,…\ldots,9999,100100。他划掉 11,跳过下一个数 (2)(2),划掉 33,并继续交替跳过和划掉直到列表末尾。然后他回到列表开头,划掉第一个剩下的数 (2)(2),跳过下一个数 (4)(4),划掉 66,跳过 88,划掉 1010,如此继续到末尾。Josh 以这种方式继续,直到只剩一个数。这个数是多少?
Josh writes the numbers 1,1, 2,2, 3,3, …,\ldots, 99,99, 100.100. He marks out 1,1, skips the next number (2),(2), marks out 3,3, and continues skipping and marking out the next number to the end of his list. Then he goes back to the start of his list, marks out the first remaining number (2),(2), skips the next number (4),(4), marks out 6,6, skips 8,8, marks out 10,10, and so on to the end. Josh continues in this manner until only one number remains. What is that number?
1313
3232
5656
6464
9696
答案:D
知识点:2的幂找规律
小提示:
第一轮后只剩 22 的倍数;下一轮后只剩 44 的倍数。
After the first pass only the multiples of 22 remain; after the next, only the multiples of 44
大提示:
每一轮留下下一个 22 的幂的倍数;找出不超过 100100 的最大 22 的幂。
Each pass keeps the multiples of the next power of 2;2; find the highest power of 22 that is at most 100100
解答:
第一轮划掉奇数,留下 22 的倍数。第二轮划掉 2,6,10,…2,6,10,\ldots 留下 44 的倍数。一般地,第 nn 轮后只剩 2n2^n 的倍数。最后幸存的是不超过 100100 的最大 22 的幂,即 26=642^6=64。
所以正确答案是 D。
The first pass removes the odd numbers, leaving the multiples of 2.2. The second pass removes 2,6,10,…,2,6,10,\ldots, leaving the multiples of 4.4. In general, after the nnth pass only the multiples of 2n2^n remain. The surviving number is the highest power of 22 not exceeding 100,100, which is 26=64.2^6=64.
Thus, the correct answer is D.
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