Difference between revisions of "2013 AIME I Problems/Problem 8"
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<math>\frac{1}{mn} \le x \le \frac{m}{n}</math> | <math>\frac{1}{mn} \le x \le \frac{m}{n}</math> | ||
| − | <math>\frac{1}{mn} - \frac{m}{n | + | <math>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</math> |
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| + | <math>n = 2013m - \frac{2013}{m}</math> | ||
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| + | For ''n'' to be an integer, ''m'' must divide 2013, and ''m'' > 1. To minimize ''n'', ''m'' should be as small as possible because increasing ''m'' will decrease <math>\frac{2013}{m}</math> , the amount you are subtracting, and increase 2013''m'' , the amount you are adding; this also leads to a small ''m'' which clearly minimizes ''m'' + ''n''. | ||
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| + | We let ''m'' equal 3, the smallest non-1 factor of 2013. <math>n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368</math> | ||
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| + | <math>m + n = 5371,</math> so the answer is <math>\boxed{371}</math> | ||
Revision as of 20:12, 15 March 2013
Problem 8
The domain of the function f(x) = arcsin(log
(nx)) is a closed interval of length 
, where m and n are positive integers and m > 1. Find the remainder when the smallest possible sum m + n is divided by 1000.
Solution
The domain of the arcsin function is [-1, 1], so -1 
log(nx) 1.
For n to be an integer, m must divide 2013, and m > 1. To minimize n, m should be as small as possible because increasing m will decrease 
, the amount you are subtracting, and increase 2013m , the amount you are adding; this also leads to a small m which clearly minimizes m + n.
We let m equal 3, the smallest non-1 factor of 2013. 
so the answer is 
