2002 AMC 12A Problems/Problem 20
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
The repeating decimal is equal to
When expressed in lowest terms, the denumerator 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 denumerator
can not be achieved, leaving us with
possible denumerators.
(The other ones are achieved e.g. for equal to
,
,
,
, and
, respectively.)
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 |