2016 AMC 12B Problems/Problem 22

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?

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

Solution

Solution by e_power_pi_times_i

If $\frac{1}{n} = 0.\overline{abcdef}$ , $n$ must be a factor of $999999$ . Also, by the same procedure, $n+6$ must be a factor of $9999$ . Checking through all the factors of $999999$ and $9999$ that are less than $1000$ , we see that $n = 297$ is a solution, so the answer is $\boxed{\textbf{(B)}}$ .

Note: $n = 27$ and $n = 3$ are both solutions, which invalidates this method. However, we need to examine all factors of $999999$ that are not factors of $99999$ , $999$ , or $99$ , or $9$ . Additionally, we need $n+6$ to be a factor of $9999$ but not $999$ , $99$ , or $9$ . Indeed, $297$ satisfies these requirements.

We can see that $n=27$ and $n=3$ are not solutions by checking it in the requirements of the problem: $\frac{1}{3}=0.3333\dots$ , period 1, and $\frac{1}{27}=0.037037\dots$ , period 3. Thus, $n=297$ is the only answer.

For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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2016 AMC 12B Problems/Problem 22

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