2015 AMC 10B Problems/Problem 16

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 :
- 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 :
- 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 $\frac9{10*9*8}=\boxed{\text{(\textbf C) }\frac1{80}}$

2015 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2015 AMC 10B Problems/Problem 16

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