Difference between revisions of "2002 AMC 12A Problems/Problem 20"
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=== Solution 3 === | === Solution 3 === | ||
− | Since <math>\frac{1}{99}=0.\overline{01}</math>, we know that <math>0.\overline{ab} = \frac{ab}{99}</math>. From here, we wish to find the number of factors of <math>99</math>, which is <math>6</math>. However, notice that <math>1</math> is not a possible denominator, so our answer is <math>6-1=\boxed{5}</math> | + | Since <math>\frac{1}{99}=0.\overline{01}</math>, we know that <math>0.\overline{ab} = \frac{ab}{99}</math>. From here, we wish to find the number of factors of <math>99</math>, which is <math>6</math>. However, notice that <math>1</math> is not a possible denominator, so our answer is <math>6-1=\boxed{5}</math>. |
+ | <cmath></cmath> | ||
+ | ~AopsUser101 | ||
== See Also == | == See Also == |
Revision as of 14:35, 6 July 2019
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
Another way to convert the decimal into a fraction (no idea what it's called). We have where are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denomenator.
~ Nafer
Solution 3
Since , we know that . From here, we wish to find the number of factors of , which is . However, notice that is not a possible denominator, so our answer is . ~AopsUser101
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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