2006 AMC 12B 第 20 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

20.

从区间 (0,1)(0, 1) 中随机选取 xx。满足 的概率是多少?这里 x\lfloor x \rfloor 表示小于或等于 xx 的最大整数。 log104xlog10x=0?\lfloor \log_{10} 4x \rfloor - \lfloor \log_{10} x \rfloor = 0?

Let xx be chosen at random from the interval (0,1).(0, 1). What is the probability that log104xlog10x=0?\lfloor \log_{10} 4x \rfloor - \lfloor \log_{10} x \rfloor = 0? Here x\lfloor x \rfloor denotes the greatest integer that is less than or equal to x.x.

18\dfrac{1}{8}

320\dfrac{3}{20}

16\dfrac{1}{6}

15\dfrac{1}{5}

14\dfrac{1}{4}

答案:C
知识点:取整函数对数几何概率
难度评级:2090
解答:

方程表示 log10x=log104x\lfloor \log_{10} x \rfloor = \lfloor \log_{10} 4x \rfloor,也就是 xx4x4x 位于同一区间 [10n,10n+1)[10^n, 10^{n+1})

这恰好在 10nx10^n \le x4x<10n+14x \lt 10^{n+1} 时成立,即 10nx<10n+1410^n \le x \lt \dfrac{10^{n+1}}{4}

[10n,10n+1)[10^n, 10^{n+1}) 内,有利部分所占比例为 10n+1/410n10n+110n=10/41101=16. \begin{aligned} &\frac{10^{n+1}/4 - 10^n}{10^{n+1} - 10^n} \\ &= \frac{10/4 - 1}{10 - 1} = \frac{1}{6}. \end{aligned}

因为这个比例在每个这样的区间上都相同,所以总概率为 16\dfrac{1}{6}

因此,正确答案是 C

The equation says log10x=log104x,\lfloor \log_{10} x \rfloor = \lfloor \log_{10} 4x \rfloor, i.e. xx and 4x4x lie in the same interval [10n,10n+1).[10^n, 10^{n+1}).

This holds exactly when 10nx10^n \le x and 4x<10n+1,4x \lt 10^{n+1}, that is 10nx<10n+14.10^n \le x \lt \dfrac{10^{n+1}}{4}.

Within [10n,10n+1),[10^n, 10^{n+1}), the favorable fraction is 10n+1/410n10n+110n=10/41101=16. \begin{aligned} &\frac{10^{n+1}/4 - 10^n}{10^{n+1} - 10^n} \\ &= \frac{10/4 - 1}{10 - 1} = \frac{1}{6}. \end{aligned}

Since this fraction is the same on every such interval, the overall probability is 16.\dfrac{1}{6}.

Thus, the correct answer is C.

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