2015 AMC 10B Problem 16

Attempt Problem 16 of the 2015 AMC 10B 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 2015 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

Al, Bill, and Cal will each randomly be assigned a whole number from 11 to 10,10, inclusive, with no two of them getting the same number. What is the probability that Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's?

91000 \dfrac{9}{1000}

190 \dfrac{1}{90}

180 \dfrac{1}{80}

172 \dfrac{1}{72}

2121 \dfrac{2}{121}

Answer: C
Concepts:basic probabilitydivisibilitysystematic listing
Difficulty rating: 1600
Solution:

Let (A,B,C)(A,B,C) be the numbers assigned to Al, Bill, and Cal. We need AA to be a multiple of BB, and BB to be a multiple of CC, with all three numbers distinct.

The valid triples are (4,2,1),(6,2,1),(8,2,1),(10,2,1),(6,3,1),(9,3,1),(8,4,1),(10,5,1),(8,4,2). \begin{gathered} (4,2,1),(6,2,1),(8,2,1), \\ (10,2,1),(6,3,1),(9,3,1), \\ (8,4,1),(10,5,1),(8,4,2). \end{gathered} There are 99 favorable assignments.

The total number of assignments is 1098=72010\cdot9\cdot8=720, so the probability is 9720=180\frac9{720}=\frac1{80}.

Thus, the correct answer is C.

← Problem 15#15
Full Exam

Problem 16 in Other Years