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.

$\frac{1}{m} \le nx \le m$

$\frac{1}{mn} \le x \le \frac{m}{n}$

$\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}$

$n = 2013m - \frac{2013}{m}$

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 m equal 3, the smallest non-1 factor of 2013. $n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$

$m + n = 5371,$ so the answer is $\boxed{371}$

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2013 AIME I Problems/Problem 8

Problem 8

Solution