2002 AMC 12A Problems/Problem 20
Contents
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 the 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 (simplifying, I guess?). We have where are digits. Continuing in the same way by looking at the factors of 99, we have 5 different possibilities for the denominator.
~ Nafer ~ edit by SpeedCuber7 ~ edit by PojoDotCom
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
Solution 4 (Alcumus)
Since , the denominator must be a factor of . The factors of are and . Since and are not both nine, the denominator cannot be . By choosing and appropriately, we can make fractions with each of the other denominators.
Thus, the answer is .
Video Solution
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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