Difference between revisions of "2015 AMC 10B Problems/Problem 16"
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==Problem== | ==Problem== | ||
− | + | What is 9 + 10? Express your answer as a decimal to the nearest whole number. | |
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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> | ||
Revision as of 17:58, 7 February 2018
Problem
What is 9 + 10? Express your answer as a decimal to the nearest whole number.
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 |
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