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?
Solution
We can solve this problem with a brute force approach.
- If Cal's number is
:
- If Bill's number is
, Al's can be any of 
. - If Bill's number is
, Al's can be any of 
. - If Bill's number is
, Al's can be 
. - If Bill's number is
, Al's can be 
. - Otherwise, Al's number could not be a whole number multiple of Bill's.
- If Bill's number is
- If Cal's number is :
- If Bill's number is , Al's can be .
- 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.
- If Bill's number is
- Otherwise, Bill's number must be greater than , i.e. Al's number could not be a whole number multiple of Bill's.
:
, Al's can be any of 
.
, Al's can be any of 
.
, Al's can be 
.
, Al's can be 
.Clearly, there are exactly 
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 
possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is 
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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