Difference between revisions of "2013 AIME I Problems/Problem 8"
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<cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}</cmath> | <cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}</cmath> | ||
− | A larger value of <math>m</math> will also result in a larger value of <math>n</math> since | + | A larger value of <math>m</math> will also result in a larger value of <math>n</math> since <math> \frac{m^2 - 1}{mn} \approx \frac{m^2}{mn}=\frac{m}{n}</math> meaning <math>m</math> and <math>n</math> increase about linearly for large <math>m</math> and <math>n</math>. So we want to find the smallest value of <math>m</math> that also results in an integer value of <math>n</math>. The problem states that <math>m > 1</math>. Thus, first we try <math>m = 2</math>: |
<cmath>\frac{3}{2n} = \frac{1}{2013} \implies 2n = 3 \cdot 2013 \implies n \notin \mathbb{Z} </cmath> | <cmath>\frac{3}{2n} = \frac{1}{2013} \implies 2n = 3 \cdot 2013 \implies n \notin \mathbb{Z} </cmath> | ||
Now, we try <math>m=3</math>: | Now, we try <math>m=3</math>: |
Revision as of 09:52, 2 September 2017
Contents
Problem 8
The domain of the function is a closed interval of length , where and are positive integers and . Find the remainder when the smallest possible sum is divided by 1000.
Solution
We know that the domain of is , so . Now we can apply the definition of logarithms: Since the domain of has length , we have that
A larger value of will also result in a larger value of since meaning and increase about linearly for large and . So we want to find the smallest value of that also results in an integer value of . The problem states that . Thus, first we try : Now, we try : Since is the smallest value of that results in an integral value, we have minimized , which is .
Solution 2
We start with the same method as above. The domain of the arcsin function is , so .
For to be an integer, must divide , and . To minimize , should be as small as possible because increasing will decrease , the amount you are subtracting, and increase , the amount you are adding; this also leads to a small which clearly minimizes .
We let equal , the smallest factor of that isn't . Then we have
, so the answer is .
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.