Question

For a certain positive integer $n$ less than $1000,$ the decimal equivalent of $\dfrac1n$ is $0.\overline{abcdef},$ a repeating decimal of period $6,$ and the decimal equivalent of $\dfrac{1}{n+6}$ is $0.\overline{wxyz},$ a repeating decimal of period $4.$ In which interval does $n$ lie?

$[1,200]$

$[201,400]$

$[401,600]$

$[601,800]$

$[801,999]$

Answer: B

Concepts:repeating decimal multiplicative order divisibility

Difficulty rating: 2270

Solution:

Period $6$ requires $n\mid10^6-1=3^3\cdot7\cdot11\cdot13\cdot37.$ Period $4$ requires $n+6\mid10^4-1=3^2\cdot11\cdot101$ but $n+6\nmid10^2-1=3^2\cdot11$ (else the period would be $1$ or $2$ ). Hence $101\mid n+6.$ 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 $1/303$ is exactly $4.$ Thus $n=297$ lies in $[201,400].$

Thus, the correct answer is B.

2016 AMC 12B Problem 22

Answer: B

Difficulty rating: 2270

Solution:

Problem 22 in Other Years