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

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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{\textbf{(B)}}</math>.
  
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Note: <math>n = 27</math> is also a solution, 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.
+
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.
  
 
For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal
 
For anyone who wants more information about repeating decimals, visit: https://en.wikipedia.org/wiki/Repeating_decimal

Revision as of 11:37, 7 January 2019

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.

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

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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