1991 AMC 12 第 28 题

先试着解答 1991 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 1991 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.

一个坛子最初装有 100100 颗黑弹珠和 100100 颗白弹珠。反复从坛中取出三颗弹珠,再按照下表用坛外的弹珠替换:
取出的弹珠 放回的弹珠
33 颗黑弹珠 11 颗黑弹珠
22 颗黑弹珠、11 颗白弹珠 11 颗黑弹珠、11 颗白弹珠
11 颗黑弹珠、22 颗白弹珠 22 颗白弹珠
33 颗白弹珠 11 颗黑弹珠、11 颗白弹珠
反复执行这一过程后,坛中可能剩下下列哪一组弹珠?
Initially an urn contains 100100 black marbles and 100100 white marbles. Repeatedly, three marbles are removed from the urn and replaced from a pile outside the urn as follows:
Marbles removed Replaced with
33 black 11 black
22 black, 11 white 11 black, 11 white
11 black, 22 white 22 white
33 white 11 black, 11 white
Which of the following sets of marbles could be the contents of the urn after repeated applications of this procedure?
22 颗黑弹珠
22 black marbles
22 颗白弹珠
22 white marbles
11 颗黑弹珠
11 black marble
11 颗黑弹珠和 11 颗白弹珠
11 black and 11 white marble
11 颗白弹珠
11 white marble
答案:B
知识点:不变量constructive process奇偶性
小提示:
白弹珠的数量每次只会改变 00 或 22
The number of white marbles always changes by 00 or 22
大提示:
在某次操作后第一次只剩至多两颗弹珠时,检查最后一次操作的可能输出
When an operation first leaves at most two marbles, inspect the possible outputs of that final operation
解答:
白弹珠数量的奇偶性始终不变,所以含一颗白弹珠的选项不可能出现。此外,最后一次操作从三颗弹珠开始时,可能留下 11 颗黑弹珠、11 颗黑弹珠和 11 颗白弹珠,或 22 颗白弹珠,但绝不会留下 22 颗黑弹珠。留下 22 颗白弹珠的状态确实可以达到:不断把 33 颗黑弹珠替换为 11 颗黑弹珠,直到剩下 22 颗黑弹珠;再两次使用“11 黑 22 白”的规则,留下 100100 颗白弹珠;然后交替使用“33 白”规则和“11 黑 22 白”规则,每两次操作把白弹珠数减少 22,直到剩下 22 颗。
因此正确答案为 B。
The parity of the number of white marbles never changes, so choices with one white marble are impossible. Also a final operation starting with three marbles can leave 11 black, 11 black and 11 white, or 22 white, but never 22 black. The state with 22 white is attainable: repeatedly replace 33 black by 11 black until 22 black remain; twice use the 11-black-22-white rule, leaving 100100 white; then alternate the 33-white rule with the 11-black-22-white rule, reducing the number of white marbles by 22 per pair until 22 remain.
Thus the correct answer is B.
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