HMMT February

Let $Q(z)$ denote the polynomial divisor. We need that the roots of $Q$ are $k$-th roots of unity. With this in mind, we might observe that solutions to $z^{7}=1$ and $z \neq 1$ are roots of $Q$, which leads to its factorization. Alternatively, we note that $$(z-1) Q(z)=z^{11}-z^{9}+z^{7}-z^{4}+z^{2}-1=\left(z^{4}-z^{2}+1\right)\left(z^{7}-1\right)$$ Solving for the roots of the first factor, $z^{2}=\frac{1+i \sqrt{3}}{2}= \pm \operatorname{cis} \pi / 3$ (we use the notation $\operatorname{cis}(x)=\cos (x)+i \sin (x))$ so that $z= \pm \operatorname{cis}( \pm \pi / 6)$. These are primitive 12 -th roots of unity. The other roots of $Q(z)$ are the primitive 7 -th roots of unity (we introduced $z=1$ by multiplication.) It follows that the answer is $\operatorname{lcm}[12,7]=84$.