Difference between revisions of "2016 AMC 12B Problems/Problem 22"

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=Problem=
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==Problem==
  
For a certain positive integer n less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\overline{abcdef}</math>, a repeating decimal of period of 6, and the decimal equivalent of <math>\frac{1}{n+6}</math> is <math>0.\overline{wxyz}</math>, a repeating decimal of period 4. In which interval does <math>n</math> lie?
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For a certain positive integer <math>n</math> less than <math>1000</math>, the decimal equivalent of <math>\frac{1}{n}</math> is <math>0.\overline{abcdef}</math>, a repeating decimal of period of <math>6</math>, and the decimal equivalent of <math>\frac{1}{n+6}</math> is <math>0.\overline{wxyz}</math>, a repeating decimal of period <math>4</math>. In which interval does <math>n</math> lie?
  
 
<math>\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]</math>
 
<math>\textbf{(A)}\ [1,200]\qquad\textbf{(B)}\ [201,400]\qquad\textbf{(C)}\ [401,600]\qquad\textbf{(D)}\ [601,800]\qquad\textbf{(E)}\ [801,999]</math>
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[[Category: Intermediate Number Theory Problems]]
  
=Solution=
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==Solution==
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
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If <math>\frac{1}{n} = 0.\overline{abcdef}</math>, <math>n</math> must be a factor of <math>999999</math>. Also, by the same procedure, <math>n+6</math> must be a factor of <math>9999</math>. Checking through all the factors of <math>999999</math> and <math>9999</math> that are less than <math>1000</math>, we see that <math>n = 297</math> is a solution, so the answer is <math>\boxed{\textbf{(B)}}</math>.
  
If <math>\frac{1}{n} = 0.\overline{abcdef}</math>, <math>n</math> must be a factor of <math>999999</math>. Also, by the same procedure, <math>n+6</math> must be a factor of <math>9999</math>. Checking through all the factors of <math>999999</math> and <math>9999</math> that are less than <math>1000</math>, we see that <math>n = 297</math> is a solution, so the answer is <math>\boxed{(B)}</math>.
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Note: <math>n = 27</math> and <math>n = 3</math> are both solutions, which invalidates this method. However, we need to examine all factors of <math>999999</math> that are not factors of <math>99999</math>, <math>999</math>, or <math>99</math>, or <math>9</math>. Additionally, we need <math>n+6</math> to be a factor of <math>9999</math> but not <math>999</math>, <math>99</math>, or <math>9</math>. Indeed, <math>297</math> satisfies these requirements.
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We can see that <math>n=27</math> and <math>n=3</math> are not solutions by checking it in the requirements of the problem: <math>\frac{1}{3}=0.3333\dots</math>, period 1, and <math>\frac{1}{27}=0.037037\dots</math>, period 3. Thus, <math>n=297</math> is the only answer.
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For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2016|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:54, 28 July 2020

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

See Also

2016 AMC 12B (ProblemsAnswer KeyResources)
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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