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Difference between revisions of "2013 AMC 12A Problems/Problem 10"

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==Solution 2==
 
==Solution 2==
 
Let us begin by working with the condition <math>0.\overline{ab} = 0.ababab\cdots,</math>. Let <math>x = 0.ababab\cdots</math>. So, <math>100x-x = ab \Rightarrow x = \frac{ab}{99}</math>. In order for this fraction <math>x</math> to be in the form <math>\frac{1}{n}</math>, <math>99</math> must be a multiple of <math>ab</math>. Hence the possibilities of <math>ab</math> are <math>1,3,9,11,33,99</math>. Checking each of these, <math>\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},</math> and <math>\frac{99}{99} = 1</math>. So the only values of <math>n</math> that have distinct <math>a</math> and <math>b</math> are <math>11,33,</math> and <math>99</math>. So, <math>11+33+99= \boxed{\textbf{(D)} 143}</math>
 
Let us begin by working with the condition <math>0.\overline{ab} = 0.ababab\cdots,</math>. Let <math>x = 0.ababab\cdots</math>. So, <math>100x-x = ab \Rightarrow x = \frac{ab}{99}</math>. In order for this fraction <math>x</math> to be in the form <math>\frac{1}{n}</math>, <math>99</math> must be a multiple of <math>ab</math>. Hence the possibilities of <math>ab</math> are <math>1,3,9,11,33,99</math>. Checking each of these, <math>\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},</math> and <math>\frac{99}{99} = 1</math>. So the only values of <math>n</math> that have distinct <math>a</math> and <math>b</math> are <math>11,33,</math> and <math>99</math>. So, <math>11+33+99= \boxed{\textbf{(D)} 143}</math>
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==Solution 3==
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Notice that we have <math>\frac{100}{n}= ab.\overline{ab}</math>
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We can subtract <math>\frac{1}{n}=00.\overline{ab}</math> to get <cmath>\frac{99}{n}=ab</cmath>
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From this we determine <math>n</math> must be a positive factor of <math>99</math>
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The factors of <math>99</math> are <math>1,3,9,11,33,</math> and <math>99</math>.
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For <math>n=1,3,</math> and <math>9</math> however, they yield <math>ab=99,33</math> and <math>11</math> which doesn't satisfy <math>a</math> and <math>b</math> being distinct.
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For <math>n=11,33</math> and <math>99</math> we have <math>ab=09,03</math> and <math>01</math>.  (Notice that <math>a</math> or <math>b</math> can be zero)
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The sum of these <math>n</math> are <math>11+33+99=143</math>
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<math>\boxed{\textbf{(D)} 143}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2013|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2013|ab=A|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:20, 29 January 2019

Problem

Let $S$ be the set of positive integers $n$ for which $\tfrac{1}{n}$ has the repeating decimal representation $0.\overline{ab} = 0.ababab\cdots,$ with $a$ and $b$ different digits. What is the sum of the elements of $S$?

$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad$

Solution 1

Note that $\frac{1}{11} = 0.\overline{09}$.

Dividing by 3 gives $\frac{1}{33} = 0.\overline{03}$, and dividing by 9 gives $\frac{1}{99} = 0.\overline{01}$.

$S = \{11, 33, 99\}$

$11 + 33 + 99 = 143$

The answer must be at least $143$, but cannot be $155$ since no $n \le 12$ other than $11$ satisfies the conditions, so the answer is $143$.

Solution 2

Let us begin by working with the condition $0.\overline{ab} = 0.ababab\cdots,$. Let $x = 0.ababab\cdots$. So, $100x-x = ab \Rightarrow x = \frac{ab}{99}$. In order for this fraction $x$ to be in the form $\frac{1}{n}$, $99$ must be a multiple of $ab$. Hence the possibilities of $ab$ are $1,3,9,11,33,99$. Checking each of these, $\frac{1}{99} = 0.\overline{01}, \frac{3}{99}=\frac{1}{33} = 0.\overline{03}, \frac{9}{99}=\frac{1}{11} = 0.\overline{09}, \frac{11}{99}=\frac{1}{9} = 0.\overline{1}, \frac{33}{99} =\frac{1}{3}= 0.\overline{3},$ and $\frac{99}{99} = 1$. So the only values of $n$ that have distinct $a$ and $b$ are $11,33,$ and $99$. So, $11+33+99= \boxed{\textbf{(D)} 143}$


Solution 3

Notice that we have $\frac{100}{n}= ab.\overline{ab}$

We can subtract $\frac{1}{n}=00.\overline{ab}$ to get \[\frac{99}{n}=ab\]

From this we determine $n$ must be a positive factor of $99$


The factors of $99$ are $1,3,9,11,33,$ and $99$.

For $n=1,3,$ and $9$ however, they yield $ab=99,33$ and $11$ which doesn't satisfy $a$ and $b$ being distinct.

For $n=11,33$ and $99$ we have $ab=09,03$ and $01$. (Notice that $a$ or $b$ can be zero)

The sum of these $n$ are $11+33+99=143$

$\boxed{\textbf{(D)} 143}$

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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