2006 AMC 12B Problem 20

Attempt Problem 20 of the 2006 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 12B solutions, or check the answer key.

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20.

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}

Answer: C
Concepts:floor and ceiling functionslogarithmgeometric probability
Difficulty rating: 2090
Solution:

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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