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2015 AMC 10B Problems/Problem 16

Revision as of 14:34, 14 July 2020 by Smarte (talk | contribs) (→‎Solution)

Problem

Al, Bill, and Cal will each randomly be assigned a whole number from $1$ to $10$, 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?

$\textbf{(A) } \dfrac{9}{1000} \qquad\textbf{(B) } \dfrac{1}{90} \qquad\textbf{(C) } \dfrac{1}{80} \qquad\textbf{(D) } \dfrac{1}{72} \qquad\textbf{(E) } \dfrac{2}{121}$

Solution

We can solve this problem with a brute force approach.

  • If Cal's number is $1$:
    • If Bill's number is $2$, Al's can be any of $4, 6, 8, 10$.
    • If Bill's number is $3$, Al's can be any of $6, 9$.
    • If Bill's number is $4$, Al's can be $8$.
    • If Bill's number is $5$, Al's can be $10$.
    • Otherwise, Al's number could not be a whole number multiple of Bill's.
  • If Cal's number is $2$:
    • If Bill's number is $4$, Al's can be $8$.
    • Otherwise, Al's number could not be a whole number multiple of Bill's while Bill's number is still a whole number multiple of Cal's.
  • Otherwise, Bill's number must be greater than $5$, i.e. Al's number could not be a whole number multiple of Bill's.

Clearly, there are exactly $9$ cases where 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. Since there are $10*9*8$ possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is $\frac9{10*9*8}=\boxed{\text{(\textbf C) }\frac1{80}}$

Give me a better strategy. This brute force strategy is pretty unwieldy. -Anonymous

See Also

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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