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, $$\begin{array}{rl}{g}_0(n+1)& =g_0(n)+g_1(n)+g_3(n)=g(n)-g_2(n),\\ g_1(n+1)& =g(n)-g_1(n),\\ g_2(n+1)& =g(n)-g_0(n),\\ g_3(n+1)& =g(n).\end{array}$$ 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, $g(n)\equiv 0(\mathrm{mod}11)$ iff $n\equiv 3(\mathrm{mod}8)$ or $n\equiv 7(\mathrm{mod}8)$, whence $g(2010)\not\equiv 0(\mathrm{mod}11)$, $g(2011)\equiv 0(\mathrm{mod}11)$.