Difference between revisions of "2013 AIME I Problems/Problem 8"
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== Solution == | == Solution == | ||
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− | <math>\frac{1}{ | + | We know that the domain of <math>\arcsin</math> is <math>[-1, 1]</math>, so <math>-1 \le \log_m nx \le 1</math>. Now we can apply the definition of logarithms: |
+ | <cmath>m^{-1} = \frac1m \le nx \le m</cmath> <cmath>\implies \frac{1}{mn} \le x \le \frac{m}{n}</cmath> | ||
+ | Since the domain of <math>f(x)</math> has length <math>\frac{1}{2013}</math>, we have that | ||
+ | <cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>\implies \frac{m^2 - 1}{mn} = \frac{1}{2013}</cmath> | ||
− | <math>\frac{1}{ | + | A larger value of <math>m</math> will also result in a larger value of <math>n</math> since the numerator increases faster than the denominator, 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> | ||
+ | Now, we try <math>m=3</math>: | ||
+ | <cmath>\frac{8}{3n} = \frac{1}{2013} \implies 3n = 8 \cdot 2013 \implies n = 8 \cdot 671 = 5368</cmath> | ||
+ | Since <math>m=3</math> is the smallest value of <math>m</math> that results in an integral <math>n</math> value, we have minimized <math>m+n</math>, which is <math>5368 + 3 = 5371 \equiv \boxed{371} \pmod{1000}</math>. | ||
− | <math>\ | + | ==Solution 2== |
+ | We start with the same method as above. The domain of the arcsin function is <math>[-1, 1]</math>, so <math>-1 \le \log_{m}(nx) \le 1</math>. | ||
− | < | + | <cmath>\frac{1}{m} \le nx \le m</cmath> <cmath>\frac{1}{mn} \le x \le \frac{m}{n}</cmath> <cmath>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</cmath> <cmath>n = 2013m - \frac{2013}{m}</cmath> |
− | For <math>n</math> to be an integer, <math>m</math> must divide <math>2013</math>, and <math>m > 1</math>. To minimize <math>n</math>, <math>m</math> should be as small as possible because increasing <math>m</math> will decrease <math>\frac{2013}{m}</math> , the amount you are subtracting, and increase <math>2013m</math> , the amount you are adding; this also leads to a small <math>n</math> which clearly minimizes <math>m+n</math>. | + | For <math>n</math> to be an integer, <math>m</math> must divide <math>2013</math>, and <math>m > 1</math>. To minimize <math>n</math>, <math>m</math> should be as small as possible because increasing <math>m</math> will decrease <math>\frac{2013}{m}</math>, the amount you are subtracting, and increase <math>2013m</math>, the amount you are adding; this also leads to a small <math>n</math> which clearly minimizes <math>m+n</math>. |
− | We let <math>m</math> equal 3, the smallest factor of <math>2013</math> that isn't <math>1</math>. Then we have <math>n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368</math> | + | We let <math>m</math> equal <math>3</math>, the smallest factor of <math>2013</math> that isn't <math>1</math>. Then we have <math>n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368</math> |
<math>m + n = 5371</math>, so the answer is <math>\boxed{371}</math>. | <math>m + n = 5371</math>, so the answer is <math>\boxed{371}</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=7|num-a=9}} | {{AIME box|year=2013|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:39, 30 December 2013
Contents
[hide]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 the numerator increases faster than the denominator, 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.