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

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== Problem 8 ==
 
== Problem 8 ==
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The domain of the function f(x) = arcsin(log<math>_{m}</math>(''nx'')) is a closed interval of length <math>\frac{1}{2013}</math> , where ''m'' and ''n'' are positive integers and ''m'' > 1. Find the remainder when the smallest possible sum ''m'' + ''n'' is divided by 1000.
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The domain of the function f(x) = arcsin(log<math>_{m}</math>(nx)) 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 ==
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The domain of the arcsin function is [-1, 1], so -1 <math>\le</math> log<math>_{m}</math>(''nx'') <math>\le</math> 1.  
+
The domain of the arcsin function is <math>[-1, 1]</math>, so <math>-1 \le</math> log_{m}(nx) \le 1<math>.  
  
−
<math>\frac{1}{m} \le nx \le m</math>
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</math>\frac{1}{m} \le nx \le m<math>
  
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<math>\frac{1}{mn} \le x \le \frac{m}{n}</math>
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</math>\frac{1}{mn} \le x \le \frac{m}{n}<math>
  
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<math>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}</math>
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</math>\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}<math>
  
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<math>n = 2013m - \frac{2013}{m}</math>
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</math>n = 2013m - \frac{2013}{m}<math>
  
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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 <math>\frac{2013}{m}</math> , the amount you are subtracting, and increase 2013''m'' , the amount you are adding; this also leads to a small ''m'' which clearly minimizes ''m'' + ''n''.
+
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>m<math> which clearly minimizes '</math>m+n<math>.
  
−
We let ''m'' equal 3, the smallest non-1 factor of 2013. <math>n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368</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>
  
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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}$.

Revision as of 18:19, 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$$ (Error compiling LaTeX. Unknown error_msg)\frac{1}{mn} \le x \le \frac{m}{n}$$ (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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