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

(Created page with "== Problem 8 == 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 posi...")
 
(Solution)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
 +
The domain of the arcsin function is [-1, 1], so -1 <math>\le</math> log<math>_{m}</math>(''nx'') <math>\le</math> 1.
 +
 +
<math>\frac{1}{m} \le nx \le m</math>
 +
 +
<math>\frac{1}{mn} \le x \le \frac{m}{n}</math>
 +
 +
<math>\frac{1}{mn} - \frac{m}{n} = \frac{1}{2013}</math>

Revision as of 21:01, 15 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.

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

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

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