Difference between revisions of "2002 AMC 12A Problems/Problem 20"
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== Solution == | == Solution == | ||
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+ | === Solution 1 === | ||
The repeating decimal <math>0.\overline{ab}</math> is equal to | The repeating decimal <math>0.\overline{ab}</math> is equal to | ||
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(The other ones are achieved e.g. for <math>ab</math> equal to <math>33</math>, <math>11</math>, <math>9</math>, <math>3</math>, and <math>1</math>, respectively.) | (The other ones are achieved e.g. for <math>ab</math> equal to <math>33</math>, <math>11</math>, <math>9</math>, <math>3</math>, and <math>1</math>, respectively.) | ||
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+ | === Solution 2 === | ||
== See Also == | == See Also == |
Revision as of 20:10, 1 July 2019
Contents
[hide]Problem
Suppose that and are digits, not both nine and not both zero, and the repeating decimal is expressed as a fraction in lowest terms. How many different denominators are possible?
Solution
Solution 1
The repeating decimal is equal to
When expressed in lowest terms, the denominator of this fraction will always be a divisor of the number . This gives us the possibilities . As and are not both nine and not both zero, the denominator can not be achieved, leaving us with possible denominators.
(The other ones are achieved e.g. for equal to , , , , and , respectively.)
Solution 2
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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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