Difference between revisions of "2013 AMC 12A Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | + | ==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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The answer must be at least <math>143</math>, but cannot be <math>155</math> since no <math>n \le 12</math> other than <math>11</math> satisfies the conditions, so the answer is <math>143</math>. | The answer must be at least <math>143</math>, but cannot be <math>155</math> since no <math>n \le 12</math> other than <math>11</math> satisfies the conditions, so the answer is <math>143</math>. | ||
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+ | ==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> less than <math>99</math>. Hence the possibilities of <math>ab</math> are <math>1,3,9,11,33</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>. 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=143 \textbf{D}</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}} |
Revision as of 19:41, 22 February 2013
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
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 less than . Hence the possibilities of are . Checking each of these, . So the only values of that have distinct and are and . So,
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 |