Part of an "n-pointed regular star" is shown. It is a simple closed polygon in which all $2n$ edges are congruent, angles $A_1,A_2,\cdots,A_n$ are congruent, and angles $B_1,B_2,\cdots,B_n$ are congruent. If the acute angle at $A_1$ is $10^\circ$ less than the acute angle at $B_1$, then $n=$
$\text{(A) } 12\quad \text{(B) } 18\quad \text{(C) } 24\quad \text{(D) } 36\quad \text{(E) } 60$
If we sum up the angles to obtain 360, we can see that the B angles add to the sum and the A angles subtract from the sum (an easy way of looking at this is by using the opposing angle theorem: if A[n] = B[n] than their total contribution is 0). Thus we have B[1] + B[2] + ... + B[n] - A[1] - A[2] - ... A[n] = 360. But every pair of A,B has a total 'angle contribution' of 10, thus there are 36 $\fbox{D}$  pairs of A,B.