HMMT November

We claim that for any choice of $x_{2}, x_{3}, \ldots, x_{2012}$, there is exactly one possible value of $x_{1}$ satisfying the condition. We have $x_{1}+2 x_{2}+\ldots+2012 x_{2012} \equiv 0(\bmod 2012)$ or $x_{1} \equiv -\left(2 x_{2}+\ldots+2012 x_{2012}\right)(\bmod 2012)$. Indeed, we see that the right hand side is always an integer between 0 and 2011, so $x_{1}$ must equal this number. Now, there are 2012 choices for each of the 2011 variables $x_{2}, \ldots, x_{2012}$, and each of the $2012^{2011}$ possible combinations gives exactly one valid solution, so the total number of 2012-tuples is $2012^{2011}$.