2002 AMC 12A Problems/Problem 20
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
Another way to convert the decimal into a fraction (no idea what it's called). We have
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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All AMC 12 Problems and Solutions |
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