smc

The repeating decimal $0.\overline{ab}$ is equal to $$\frac{10a+b}{100}+\frac{10a+b}{10000}+\cdots =(10a+b)\cdot \left(\frac{1}{10^2}+\frac{1}{10^4}+\cdots \right)=(10a+b)\cdot \frac{1}{99}=\frac{10a+b}{99}$$ When expressed in the lowest terms, the denominator of this fraction will always be a divisor of the number $99=3\cdot 3\cdot 11$. This gives us the possibilities $\{1,3,9,11,33,99\}$. As $a$ and $b$ are not both nine and not both zero, the denominator $1$ can not be achieved, leaving us with $\boxed{5}$ possible denominators. (The other ones are achieved e.g. for $ab$ equal to $33$, $11$, $9$, $3$, and $1$, respectively.)