Let $p_n$ denote the total number of triangles formed by joining the vertices of an $n$-sided regular polygon. If $p_{n+1} - p_n = 66$,then the sum of all distinct prime divisors of $n$ is:

$A$ regular polygon has $170$ diagonals. Then the measure of the interior angle of the polygon is

There are $12$ points in a plane,no three of which are in the same straight line,except $5$ points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these $12$ points is

There are $m$ points on a straight line $AB$ and $n$ points on another line $AC$,none of them being the point $A$. Triangles are formed from these points as vertices when $(i)$ $A$ is excluded $(ii)$ $A$ is included. Then the ratio of the number of triangles in the two cases is

$l_1, l_2, l_3$ are three parallel lines in the same plane. If there are $m$ points on $l_1$,$n$ points on $l_2$,and $k$ points on $l_3$,what is the maximum number of triangles that can be formed with vertices at these points?

In the given figure,$6$ '$A$'s should be written in such a manner that every row contains at least one '$A$'. The number of ways this is possible is