# 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 $1$:
• 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 $\frac9{10*9*8}=\boxed{\text{(\textbf C) }\frac1{80}}$

## See Also

 2015 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.