smc

Each angle in each equiangular polygon is $\frac{180(n-2)}{n}=180-\frac{360}{n}$, where $n$ is the number of sides. As $n$ gets larger, each angle in the equiangular polygon approaches (but does not reach) $180^{\circ }$. That means $kx<180$, so $x<\frac{180}{k}$. Recall that each angle in an equiangular triangle has $60^{\circ }$, each angle in an equiangular quadrilateral has $90^{\circ }$, and so on. If $k=2$, then $x<90$. That means the only option available is $x=60$ because an equiangular triangle is the only equiangular polygon to have acute angles. If $k=3$, then $x<60$, and there are no values of $x$ that satisfy. As $k$ increases, $\frac{180}{k}$ decreases even further, so we can for sure say that there is only 1 possibility for the pair $(x,k)$, which is answer choice $\boxed{1}$.