Romanian Mathematical Olympiad

a) $x_1 = 1$ (0 is divisible with $4$), $x_2 = 4$ (the numbers $12$, $16$, $20$ and $60$ are divisible with $4$), $x_3 = 3 \cdot 5$, (because the first digit cannot be $0$ and the last two can be $12$, $16$, $20$, $60$ and $00$), $x_4 = 3 \cdot 4 \cdot 5 = 60$ (because the first digit cannot be $0$, for the second digit there are $4$ possibilities and the last two digits can be $12$, $16$, $20$, $60$ and $00$).

b) If $n \ge 3$, then a number $A$ which fulfills the hypothesis is of the form
$$A = \overline{a_1 a_2 \dots a_{n-2} p q},$$
where its first digit can have three values, each of the digits $a_2$, $a_3$, $\dots$, $a_{n-2}$ can be chosen in $4$ ways and the last two digits can be $12$, $16$, $20$, $60$ or $00$. So $x_n = 3 \cdot 4^{n-3} \cdot 5$, for every $n \ge 3$.
For $n \ge 3$, $\frac{x_{n+1}}{x_n} = 4$, whence $1 + \left\lfloor \frac{x_2}{x_1} \right\rfloor + \left\lfloor \frac{x_3}{x_2} \right\rfloor + \left\lfloor \frac{x_4}{x_3} \right\rfloor + \dots + \left\lfloor \frac{x_{n+1}}{x_n} \right\rfloor = 1 + 4 + 3 + 4(n-2)$, $4n = 2016$, $n = 504$.