2013 AIME I Problems/Problem 8

Problem 8

The domain of the function f(x) = arcsin(log $_{m}$ (nx)) is a closed interval of length $\frac{1}{2013}$ , 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 $-1 \le$ log_{m}(nx) \le 1 $.$ \frac{1}{m} \le nx \le m$$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{mn} \le x \le \frac{m}{n}$$ (Error compiling LaTeX. Unknown error_msg)\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}$$ (Error compiling LaTeX. Unknown error_msg)n = 2013m - \frac{2013}{m} $For$ 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 $will decrease$ \frac{2013}{m} $, the amount you are subtracting, and increase$ 2013m $, the amount you are adding; this also leads to a small$ m $which clearly minimizes '$ m+n$.

We let$ (Error compiling LaTeX. Unknown error_msg)m $equal 3, the smallest factor of$ 2013 $that isn't$ 1 $..$ n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$$ (Error compiling LaTeX. Unknown error_msg)m + n = 5371 $, so the answer is$ \boxed{371}$.

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2013 AIME I Problems/Problem 8

Problem 8

Solution