Difference between revisions of "2016 AMC 12B Problems/Problem 22"

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For a certain positive integer n less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\overline{abcdef}</math>, a repeating decimal of period of 6, and the decimal equivalent of <math>\frac{1}{n+6}</math> is <math>0.\overline{wxyz}</math>, a repeating decimal of period 4. In which interval does <math>n</math> lie?
 
For a certain positive integer n less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\overline{abcdef}</math>, a repeating decimal of period of 6, and the decimal equivalent of <math>\frac{1}{n+6}</math> is <math>0.\overline{wxyz}</math>, a repeating decimal of period 4. In which interval does <math>n</math> lie?
  
<math>\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]1\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]</math>
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<math>\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]</math>

Revision as of 11:54, 21 February 2016

Problem

For a certain positive integer n less than $1000$, the decimal equivalent of $\frac{1}{n}$ is $0.\overline{abcdef}$, a repeating decimal of period of 6, and the decimal equivalent of $\frac{1}{n+6}$ is $0.\overline{wxyz}$, a repeating decimal of period 4. In which interval does $n$ lie?

$\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]$