2017 AMC 12B Problem 20

Attempt Problem 20 of the 2017 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 2017 AMC 12B solutions, or check the answer key.

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20.

Real numbers xx and yy are chosen independently and uniformly at random from the interval (0,1).(0, 1). What is the probability that log2x=log2y,\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor, where r\lfloor r \rfloor denotes the greatest integer less than or equal to the real number r?r?

18\dfrac{1}{8}

16\dfrac{1}{6}

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

Answer: D
Concepts:geometric probabilityfloor and ceiling functionsgeometric sequence
Difficulty rating: 1990
Solution:

For each positive integer n,n, log2x=n\lfloor \log_2 x \rfloor = -n exactly when 12nx<12n1,\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}}, an interval of length 12n.\dfrac{1}{2^n}. The event that both floors equal n-n is a square of area 14n.\dfrac{1}{4^n}. Summing over all n,n, the probability is n=114n=1/411/4=13.\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{1/4}{1 - 1/4} = \frac{1}{3}.

Thus, the correct answer is D.

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