2017 AMC 12B Problem 20

Below is the professionally curated solution for Problem 20 of the 2017 AMC 12B, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:geometric probabilityfloor and ceiling functionsgeometric sequence

Difficulty rating: 1990

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}

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.

Problem 20 in Other Years