67th 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 $$A=\overline{a_1{a}_2\dots a_{n-2}pq},$$ 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 $1+\lfloor \frac{x_2}{x_1}\rfloor +\lfloor \frac{x_3}{x_2}\rfloor +\lfloor \frac{x_4}{x_3}\rfloor +\cdots +\lfloor \frac{x_{n+1}}{x_n}\rfloor =1+4+3+4(n-2)$, $4n=2016$, $n=504$.