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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

Solution 2 (Faster Approach)

Notice that the repeating fraction $0.\overline{abcdef}$ can be represented as $\frac{abcdef}{999999},$ and thereby, $n|999999.$ Also, notice that $0.\overline{wxyz} = \frac{wxyz}{9999},$ so $(n+6)|9999.$ However, we have to make some restrictions here. For instance, if $n|99999,$ then $\frac{1}{n}$ could be expressed as $\frac{a’b’c’d’e’}{99999} = .\overline{a’b’c’d’e’}$ which cannot happen. Therefore, from this, we see that the smallest $m$ such that $n|\underbrace{999\cdots999}_{m \text{ nines}}$ is $m = 6.$ Also, the smallest number $m$ such that $(n+6)|\underbrace{999\cdots999}_{m \text{ nines}}$ is $m = 4$ by similar reasoning.

Proceeding, we can factorize $9999 = 99 \times 101,$ after which we see that $n+6$ must contain a prime factor of $101$ as it cannot divide $99$ but must divide $9999.$ However, $101$ is prime, so $101|(n+6)$! Looking at the answer choices, all of the intervals are less than $1000,$ so we know that (the minimum value of) $n+6$ is thereby either $101, 101 \times 3,$ or $101 \times 9.$ Testing, we see that $n+6 = 303$ gives $n = 297 = 3^3 \times 11,$ which in fact is a divisor of $999 \times 1001$ while not being a divisor of $999.$ Therefore, the answer is $\boxed{\text{(B)}}.$

~ Professor-Mom

Video Solution by CanadaMath (Problem 21-25)

Fast Forward to 11:20 for problem 22 https://www.youtube.com/watch?v=P3jJDLGyF2w&t=1546s

~THEMATHCANADIAN

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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