Belarus — Selection and Training Session

Let $g_0(n)$, $g_1(n)$, $g_2(n)$, $g_3(n)$ be the quantities of the numbers satisfying the given condition which end by the digits $0$, $1$, $2$, $3$ respectively. Then $g(n) = g_0(n) + g_1(n) + g_2(n) + g_3(n)$. By condition,
$$\begin{align*} g_0(n+1) &= g_0(n) + g_1(n) + g_3(n) = g(n) - g_2(n), \\ g_1(n+1) &= g(n) - g_1(n), \\ g_2(n+1) &= g(n) - g_0(n), \\ g_3(n+1) &= g(n). \end{align*}$$
Summing all the equalities, we get
$$g(n+1) = 3g(n) + g_3(n) = 3g(n) + g(n-1).$$
Note that modulo $11$ we have $g(1) \equiv 3$, $g(2) \equiv 10$, $g(3) \equiv 0$, $g(4) \equiv 10$, $g(5) \equiv 8$, $g(6) \equiv 1$, $g(7) \equiv 0$, $g(8) \equiv 1$, $g(9) \equiv 3$, $g(10) \equiv 10$. Now we see that residues of the numbers $g(n)$ modulo $11$ repeat periodically with period $8$. So, $g(n) \equiv 0 \pmod{11}$ iff $n \equiv 3 \pmod{8}$ or $n \equiv 7 \pmod{8}$, whence $g(2010) \not\equiv 0 \pmod{11}$, $g(2011) \equiv 0 \pmod{11}$.