What is the least possible denominator of such a positive rational number? In a contest, I failed this problem. Can anyone give me an hint or show me how to solve the problem?

What is the least possible denominator of a positive rational number whose repeating decimal representation is ， where $A$ and $B$ are distinct digits?

 A: You can just go through the numbers: 


*

*Denominators of $3,6,9$ only contain single-digit repeats

*Denominators of $2,4,5,8,10$ have no repeats

*Denominators of $7$ repeat in groups of $6$
$11$ works.
A: Let $n$ be the natural number with representation $AB$ in base $10$. For such a number $x$, one has $100x=n+x$, whence $\;x=\dfrac n{99}$. This shows the irreducible fraction representing $x$ has a divisor of $99$ as denominator.
The fraction cannot be simplified by $11$, for this would imply $A=B$ (divisibility  by $11$ criterion). But it may be simplified by $9$ if $A+B=9$ (e.g. $\;0.\overline{18}$). Hence the smallest denominator is equal to $11$, and it is obtained if $n$ is divisible by $9$.
A: You've probably see this old trick, but it's worth going through.
Since we are only interested in the fractional part of the number, we need only look at the equation
$$\dfrac ND = 0.\overline{ab} = 0.abababab\dots$$
where
$\qquad 0\le a,b \le 9$,
$\qquad ab\; \text{represents the two-digit number $10a+b$}$.
$\qquad 0 < N < D$.
We compute
\begin{array}{rll}
   100\, \dfrac ND &=& ab.abababab\dots \\
    -\dfrac     ND &=& -0.abababab\dots \\
\hline
    99\, \dfrac ND &=& ab \\
\end{array}
So $\dfrac ND = \dfrac{ab}{99}$
It's possible that the fraction $\dfrac{ab}{99}$ can be reduced to a fraction with a smaller denominator. That would involve the two-digit number $ab$ being a multiple of $3, 9, 11$ or $33$, since the only proper divisors of $99$ are $3, 9, 11$ and $33$. The corresponding denominators would therefore be
$33, 11, 9$ and $3$.
Denominators of $3$ and $9$ produce fractions with $a$ and $b$ being the same digit. For example $\dfrac 23 = 0.\overline{66}$ and $\dfrac 59 = 0.\overline{55}$.
Denominators of $11$ will produce two-digit repitends {repeating parts}.
for example
$\dfrac{7}{11} = \dfrac{63}{99} = 0.\overline{63}$.
So the smallest denominator is $11$.
