Difference between revisions of "2015 AMC 10B Problems/Problem 16"
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<math>\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} </math> | <math>\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} </math> | ||
− | ==Video | + | ==Video Solution== |
https://www.youtube.com/watch?v=vulB2z_PdRE&feature=youtu.be | https://www.youtube.com/watch?v=vulB2z_PdRE&feature=youtu.be |
Latest revision as of 20:24, 26 December 2020
Contents
Problem
Al, Bill, and Cal will each randomly be assigned a whole number from to , 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?
Video Solution
https://www.youtube.com/watch?v=vulB2z_PdRE&feature=youtu.be
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 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.
- Otherwise, Bill's number must be greater than , i.e. Al's number could not be a whole number multiple of Bill's.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.