2005 AMC 8 第 24 题

先试着解答 2005 AMC 8 第 24 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 2005 AMC 8 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。

24.

某计算器只有两个键 [+1+1] 和 [×2\times 2]。按下其中一个键时,计算器会自动显示结果。例如,如果计算器原来显示“99”,按 [+1+1] 后会显示“1010”。如果接着按 [×2\times 2],会显示“2020”。从显示“11”开始,要达到“200200”最少需要按多少次键?
A certain calculator has only two keys [+1+1] and [×2\times 2]. When you press one of the keys, the calculator automatically displays the result. For instance, if the calculator originally displayed “99” and you pressed [+1+1], it would display “10.10.” If you then pressed [×2\times 2], it would display “20.20.” Starting with the display “1,1,” what is the fewest number of keystrokes you would need to reach “200200”?
88
99
1010
1111
1212
答案:B
知识点:逆推法最优化
小提示:
从 200200 反向推回 11
Work backward from 200200 to 11.
大提示:
为了给出下界,注意八次按键在 200200 附近能到达和不能到达什么
For a lower bound, note what eight keystrokes can and cannot reach near 200200.
解答:
从 200200 反向操作。数为偶数时,用除以 22 撤销 ×2\times2;数为奇数时,用减去 11 撤销 +1+1。
200,100,50,25,24,12,6,3,2,1200,100,50,25,24,12,6,3,2,1 使用 99 个反向步骤,所以 99 次按键足够。
对于目标 200200,88 次按键不够:小于目标的最大可达值由先 ×2,+1\times2,+1,再六次加倍得到,即 3⋅26=1923\cdot2^6=192。更大的可能会至少达到 256256,所以 200200 不能在 88 次按键内到达。
所以正确答案是 B。
Work backward from 200200. When the number is even, undo ×2\times2 by dividing by 22; when it is odd, undo +1+1 by subtracting 11.
200,100,50,25,24,12,6,3,2,1200,100,50,25,24,12,6,3,2,1 uses 99 reverse steps, so 99 keystrokes are enough.
Eight keystrokes are not enough: the largest value below 200200 reachable in 88 keystrokes is obtained by ×2,+1\times2,+1 followed by six doublings, giving 3⋅26=1923\cdot2^6=192. The next larger possibilities overshoot to at least 256256, so 200200 cannot be reached in 88 keystrokes.
Thus, B is the correct answer.
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