<s>[INST] Repeating decimals 
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 $6$, 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] $

All I have for this is $n \mid 10^6 - 1$ and $n+ 6 \mid 10^4 - 1$. How do I proceed?
 [/INST] You have $n+6 | 9 \cdot 11 \cdot101$
But if $n+6 | 10^a -1 $ where $a<4 $, then $\frac {1}{n+6} $will have $a$ repeating digits. So checking cases we get 
$n+6 = 101, 303 $ or $909$
Now we have $n | 11 \cdot 13 \cdot 7 \cdot27 \cdot37$
So $ n= 297$  is the only solution 
EDIT:
$ \frac{1}{297} =0. \overline{003367}$
$ \frac {1}{303} =0. \overline{0033}$
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