Difference between revisions of "2013 AIME I Problems/Problem 8"
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− | == Problem | + | == Problem == |
The domain of the function <math>f(x) = \arcsin(\log_{m}(nx))</math> is a closed interval of length <math>\frac{1}{2013}</math> , where <math>m</math> and <math>n</math> are positive integers and <math>m>1</math>. Find the remainder when the smallest possible sum <math>m+n</math> is divided by 1000. | The domain of the function <math>f(x) = \arcsin(\log_{m}(nx))</math> is a closed interval of length <math>\frac{1}{2013}</math> , where <math>m</math> and <math>n</math> are positive integers and <math>m>1</math>. Find the remainder when the smallest possible sum <math>m+n</math> is divided by 1000. | ||
− | == Solution == | + | == Solution 1== |
We know that the domain of <math>\text{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: | We know that the domain of <math>\text{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: | ||
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<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>. | ||
− | ==Operation Quadratics | + | ==Solution 3 (Operation Quadratics)== |
Note that we need <math>-1\le f(x)\le 1</math>, and this eventually gets to <math>\frac{m^2-1}{mn}=\frac{1}{2013}</math>. From there, break out the quadratic formula and note that <cmath>m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.</cmath> Then we realize that the square root, call it <math>a</math>, must be an integer. Then <math>(a-n)(a+n)=4026^2.</math> | Note that we need <math>-1\le f(x)\le 1</math>, and this eventually gets to <math>\frac{m^2-1}{mn}=\frac{1}{2013}</math>. From there, break out the quadratic formula and note that <cmath>m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.</cmath> Then we realize that the square root, call it <math>a</math>, must be an integer. Then <math>(a-n)(a+n)=4026^2.</math> | ||
Latest revision as of 19:07, 24 January 2021
Problem
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 1
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 .
Solution 3 (Operation Quadratics)
Note that we need , and this eventually gets to . From there, break out the quadratic formula and note that Then we realize that the square root, call it , must be an integer. Then
Observe carefully that ! It is not difficult to see that to minimize the sum, we want to minimize as much as possible. Seeing that is even, we note that a belongs in each factor. Now, since we want to minimize to minimize , we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of and fails; the next best is and , in which and . That is our best solution, upon which we see that , thus .
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.