Problem 2. For any non-zero natural number $n$, we denote by $x_{n}$ the number of natural numbers with $n$ digits, divisible by 4, formed using the digits 2, 0, 1, or 6.

a) Calculate $x_{1}, x_{2}, x_{3}$, and $x_{4}$.

b) Find the natural number $n$ such that

$$
1+\left[\frac{x_{2}}{x_{1}}\right]+\left[\frac{x_{3}}{x_{2}}\right]+\left[\frac{x_{4}}{x_{3}}\right]+\ldots+\left[\frac{x_{n+1}}{x_{n}}\right]=2016
$$

where $[a]$ represents the integer part of the real number $a$.
Solution. a) $x_{1}=1$ (0 is divisible by 4), $x_{2}=4$ (the numbers 12, 16, 20 and 60 are divisible by 4), $x_{3}=3 \cdot 5$, (the first digit cannot be 0, and the last two can be 12, 16, 20, 60, and 00), $x_{4}=3 \cdot 4 \cdot 5=60$ (the first digit cannot be 0, the second has 4 possibilities, and the last two can be
b) If $n \geq 3$, a number $A$ that satisfies the conditions in the statement is of the form $A=\overline{a_{1} a_{2} \ldots a_{n-2} p q}$ where the first digit can take 3 values, each of the digits $a_{2}, a_{3}, \ldots, a_{n-2}$ can be chosen in 4 ways, and the last two can be 12, $16,20,60$ and 00.

It follows that $x_{n}=3 \cdot 4^{n-3} \cdot 5$ for any $n \geq 3 \ldots \ldots \ldots \ldots \ldots 2$ points

For any $n \geq 3, \frac{x_{n+1}}{x_{n}}=4$, hence $1+\left[\frac{x_{2}}{x_{1}}\right]+\left[\frac{x_{3}}{x_{2}}\right]+\left[\frac{x_{4}}{x_{3}}\right]+\ldots+$ $\left[\frac{x_{n+1}}{x_{n}}\right]=1+4+3+4(n-2), 4 n=2016, n=504 \ldots \ldots \ldots \ldots .2$ points.