Let $$X$$ be the set of all even positive integers $$n$$ such that the measure of the angle of some regular polygon is $$n$$ degrees. Find the number of elements in $$X$$.

For a regular polygon with $$k$$ sides $$\bigl(k \ge 3,\,k\in\mathbb{N}\bigr)$$ the measure of each interior angle is

$$\theta \;=\;180^\circ-\frac{360^\circ}{k}\quad -(1)$$

The question asks for all even positive integers $$n$$ that can equal $$\theta$$. Put $$\theta=n$$ in (1):

$$n \;=\;180-\frac{360}{k}\;\Longrightarrow\;\frac{360}{k}=180-n\quad -(2)$$

Let $$d=180-n$$. Then $$d=\dfrac{360}{k}$$, so $$d$$ must be a positive divisor of $$360$$. Also, because the polygon is convex, $$n\lt 180$$, hence $$d=180-n\gt 0$$.

From (2) we have $$k=\dfrac{360}{d}$$. The requirement $$k\ge 3$$ gives $$\dfrac{360}{d}\ge 3\;\Longrightarrow\;d\le 120\quad -(3)$$

Finally, the problem restricts $$n$$ to be even. Since $$180$$ is even, $$n=180-d$$ is even ⇔ $$d$$ is even. Therefore:

• $$d$$ is an even positive divisor of $$360$$.
• $$d\le 120$$ from (3).

List the divisors of $$360$$ (using $$360=2^{3}\!\cdot\!3^{2}\!\cdot\!5$$) and keep those that satisfy both conditions:

Even divisors of $$360$$ not exceeding $$120$$:
$$2,\,4,\,6,\,8,\,10,\,12,\,18,\,20,\,24,\,30,\,36,\,40,\,60,\,72,\,90,\,120$$

Total count = 16.

Each such $$d$$ gives a unique even interior angle $$n=180-d$$ in the open range $$(0,180)$$, so the set $$X$$ has exactly 16 elements.

Hence, the required number of even interior angles is 16.