Regular polygons $mod \ 10$

so required is 2x, 2x=180-360/n

180-360/n=10k => n=36/(18-k)

factors of 36 are 1,2,3,4,6,9,12,36

for 18-k,36 is not permissible as n>2

so other than 36 there 7 factors

If an interior angle of a regular polygon is a multiple of $10$ , then the exterior angle of a regular polygon must also be a multiple of $10$ (since both add up to $180°$ , another multiple of $10$ ).

Since each exterior angle is $\frac{360}{n}$ , the number of angle solutions that are a multiple of $10$ would be the number of factors of $\frac{360}{10} = 36$ that are greater or equal to $3$ (since a polygon must have at least $3$ sides). Since $36 = 2^23^2$ , its total number of factors is $(2 + 1)(2 + 1) = 9$ , and excluding $1$ and $2$ since a polygon must have at least $3$ sides, the total number of polygons such that their angles (in degrees) are a multiple of $10$ is $9 - 2 = \boxed{7}$ .