Power Problem....

Factor out $a^{2012}$ to get the expression $a^{2012}(a^1+a^2+a^3+\ldots+a^{1000})$ . Now, the expression inside the parentheses can be simplified by using the formula for the partial sum of a geometric series, $\frac{a_0(1-r^n)}{1-r}$ , so with $a_0=1$ and $r=a$ , we get $a^{2012}(\frac{1-a^{1001}}{1-a})$ . Now, this quantity will always be an odd integer, because both the numerator and the denominator are odd integers (as $a$ is even). Therefore, the largest power of two that divides this sum is $\boxed{2012}$ , because the factored out term $a^{2012}$ will have $2^{2012}$ as one of its factors.