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

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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>.
  
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We let <math>m</math> equal 3, the smallest factor of <math>2013</math> that isn't <math>1</math>.. <math>n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368</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>
  
 
<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>.

Revision as of 18:21, 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 $\boxed{371}$.

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