3. The difference of the reciprocal values of two integers is equal to $0.0 a a \ldots$, where $a \in\{1,2, \ldots, 9\}$. For which values of $a$ does the problem have solutions and how many?
Solution. We need to determine an integer $x(x \neq 0,-1)$ such that

$$
\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x(x+1)}=0.0 a a \ldots=\frac{a}{90}
$$

i.e., $\quad x(x+1)=\frac{90}{a}$, where $a \in\{1,2, \ldots, 9\}$. The right-hand side of the last equation takes integer values $90,45,30,18,15,10$ for the respective values of $a$: $1,2,3,5,6,9$. However, of these six numbers, only 90 and 30 can be written as the product of two consecutive integers:

$$
90=9 \cdot 10=-10 \cdot(-9), \quad 30=5 \cdot 6=-6 \cdot(-5)
$$

Thus, the desired values for $a$ are 1 and 3 (the corresponding values for $x$ are 9 and -10, respectively, or 5 and -6).