2016 AMC 12B Problem 22

Period $6$ means $10^6-1=3^3\cdot7\cdot11\cdot13\cdot37$ is divisible by $n;$ period $4$ means $10^4-1=3^2\cdot11\cdot101$ is divisible by $n+6$

$n+6$ must divide $10^4-1$ but not $10^2-1,$ forcing $n+6=101k;$ test each $n=101k-6\lt1000$ to see whether $10^6-1$ is divisible by $n$

Period $6$ requires $10^6-1=3^3\cdot7\cdot11\cdot13\cdot37$ to be divisible by $n.$ Period $4$ requires $10^4-1=3^2\cdot11\cdot101$ to be divisible by $n+6,$ while $10^2-1=3^2\cdot11$ is not divisible by $n+6$ (else the period would be $1$ or $2$). Hence $n+6$ is a multiple of $101.$ Since $n+6$ also divides $3^2\cdot11\cdot101$ and is less than $1006,$ the only possibilities are $n+6=101,303,909,$ giving $n=95,297,903.$ Only $297=3^3\cdot11$ divides $10^6-1,$ so $n=297.$

Finally, $10^6\equiv1\pmod{297},$ while $10^2\not\equiv1$ and $10^3\not\equiv1\pmod{297},$ so its period is exactly $6.$ Also $303$ divides $10^4-1$ but not $10^2-1,$ so the period of $\frac{1}{303}$ is exactly $4.$ Thus $n=297$ lies in $[201,400].$

Thus, the correct answer is B.