Difference between revisions of "2013 AIME I Problems/Problem 8"

Revision as of 20:44, 16 March 2013

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 $n$ which clearly minimizes $m+n$ .

We let $m$ equal 3, the smallest factor of $2013$ that isn't $1$ . Then we have $n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$

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

@@ Line 18: / Line 18: @@
 <math>m + n = 5371</math>, so the answer is <math>\boxed{371}</math>.
+== See also ==
+{{AIME box|year=2013|n=I|num-b=7|num-a=9}}

2013 AIME I (Problems • Answer Key • Resources)
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Difference between revisions of "2013 AIME I Problems/Problem 8"

Revision as of 20:44, 16 March 2013

Problem 8

Solution

See also