2016 AMC 12B Problem 22

Below is the professionally curated solution for Problem 22 of the 2016 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12B solutions, or check the answer key.

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Concepts:repeating decimal multiplicative order divisibility

Difficulty rating: 2270

22.

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?

$[1,200]$

$[201,400]$

$[401,600]$

$[601,800]$

$[801,999]$

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 $101\mid n+6,$ so $n=101k-6.$ For $n\lt1000,$ $k\in\{1,3,9\},$ giving $n\in\{95,297,903\}.$ Only $297=3^3\cdot11$ divides $10^6-1,$ so $n=297,$ which lies in $[201,400].$

Thus, the correct answer is B.

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