Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Problem)
(→‎Video Solution by CanadaMath (Problem 21-25))
 
(13 intermediate revisions by 8 users not shown)
Line 1: Line 1:
−
=Problem=
+
==Problem==
  
 
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?
 
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>
 +
[[Category: Intermediate Number Theory Problems]]
  
−
=Solution=
+
==Solution==
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
  
 +
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>.
 +
 +
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.
 +
 +
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.
 +
 +
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 <math>0.\overline{abcdef}</math> can be represented as <math>\frac{abcdef}{999999},</math> and thereby, <math>n|999999.</math> Also, notice that <math>0.\overline{wxyz} = \frac{wxyz}{9999},</math> so <math>(n+6)|9999.</math> However, we have to make some restrictions here. For instance, if <math>n|99999,</math> then <math>\frac{1}{n}</math> could be expressed as <math>\frac{a’b’c’d’e’}{99999} = .\overline{a’b’c’d’e’}</math> which cannot happen. Therefore, from this, we see that the smallest <math>m</math> such that <math>n|\underbrace{999\cdots999}_{m \text{ nines}}</math> is <math>m = 6.</math> Also, the smallest number <math>m</math> such that <math>(n+6)|\underbrace{999\cdots999}_{m \text{ nines}}</math> is <math>m = 4</math> by similar reasoning.
 +
 +
Proceeding, we can factorize <math>9999 = 99 \times 101,</math> after which we see that <math>n+6</math> must contain a prime factor of <math>101</math> as it cannot divide <math>99</math> but must divide <math>9999.</math> However, <math>101</math> is prime, so <math>101|(n+6)</math>! Looking at the answer choices, all of the intervals are less than <math>1000,</math> so we know that (the minimum value of) <math>n+6</math> is thereby either <math>101, 101 \times 3,</math> or <math>101 \times 9.</math> Testing, we see that <math>n+6 = 303</math> gives <math>n = 297 = 3^3 \times 11,</math> which in fact is a divisor of <math>999 \times 1001</math> while not being a divisor of <math>999.</math> Therefore, the answer is <math>\boxed{\text{(B)}}.</math>
 +
 +
~ 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
  
−
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>.
 
 
==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}}

Latest revision as of 23:33, 9 November 2024

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_22&oldid=232942"