Difference between revisions of "2013 AMC 12A Problems/Problem 10"
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<math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad </math> | <math> \textbf{(A)}\ 11\qquad\textbf{(B)}\ 44\qquad\textbf{(C)}\ 110\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 155\qquad </math> | ||
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==Solution 1== | ==Solution 1== | ||
Note that <math>\frac{1}{11} = 0.\overline{09}</math>. | Note that <math>\frac{1}{11} = 0.\overline{09}</math>. | ||
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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== | ||
+ | |||
+ | Notice that we have <math>\frac{100}{n}= ab.\overline{ab}</math> | ||
+ | |||
+ | We can subtract <math>\frac{1}{n}=00.\overline{ab}</math> to get <cmath>\frac{99}{n}=ab</cmath> | ||
+ | |||
+ | From this we determine <math>n</math> must be a positive factor of <math>99</math> | ||
+ | |||
+ | |||
+ | The factors of <math>99</math> are <math>1,3,9,11,33,</math> and <math>99</math>. | ||
+ | |||
+ | 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) | ||
+ | |||
+ | The sum of these <math>n</math> are <math>11+33+99=143</math> | ||
+ | |||
+ | <math>\boxed{\textbf{(D)} 143}</math> | ||
+ | |||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | As in previous solutions, we have <math>n|99</math> and <math>\overline{ab} = 99/n</math>. If we had <math>a=b</math>, the decimal would be <math>0.\overline{a}</math>, which is characterized by <math>n|9</math> and <math>a = 9/n</math>. So we seek the sum of the factors of 99 that are not also factors of 9. | ||
+ | |||
+ | Since <math>99 = 3^2 \cdot 11</math>, the sum is <math>(1 + 3 + 9)(1 + 11) - (1 + 3 + 9) = 13(12 - 1) = \textbf{(D)} 143</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=XQpQaomC2tA | ||
+ | |||
+ | ~sugar_rush | ||
== 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}} |
Latest revision as of 18:38, 30 April 2021
Problem
Let be the set of positive integers for which has the repeating decimal representation with and different digits. What is the sum of the elements of ?
Solution 1
Note that .
Dividing by 3 gives , and dividing by 9 gives .
The answer must be at least , but cannot be since no other than satisfies the conditions, so the answer is .
Solution 2
Let us begin by working with the condition . Let . So, . In order for this fraction to be in the form , must be a multiple of . Hence the possibilities of are . Checking each of these, and . So the only values of that have distinct and are and . So,
Solution 3
Notice that we have
We can subtract to get
From this we determine must be a positive factor of
The factors of are and .
For and however, they yield and which doesn't satisfy and being distinct.
For and we have and . (Notice that or can be zero)
The sum of these are
Solution 4
As in previous solutions, we have and . If we had , the decimal would be , which is characterized by and . So we seek the sum of the factors of 99 that are not also factors of 9.
Since , the sum is .
Video Solution
https://www.youtube.com/watch?v=XQpQaomC2tA
~sugar_rush
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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