Suppose that there are n points in the plane and that each of these
points is joined to exactly three of the others by straight line segments.
Then a possible value of n would be	Options:
A) 11
B) 15
C) 18
D) 27
E) 39
Count the number of pairs (p,l) where p is one of the n points and l is a line   segment joining p to one of the other n-1 points. Each point is joined to exactly 3 other   points by line segments so that the total number of pairs is 3n (three line segments for   each of the n points). On the other hand, if there are m total line segments, then each line   segment joins exactly two points so that the total number of pairs (p,l) is 2m (two pointsfor each of the m line segments). Thus, we must have 3n = 2m. But this means that n   must be even. The only even choice given is 18, so this must be the answer.   Is n = 18 really possible or is there some other argument that shows n must be divisible   by 23? One can justify that in fact if n is even, then one can join each of any n points   to 3 of the remaining n - 1 points by line segments. To see this, let n = 2k, and let the   points be p1,..., pn. For each j, join pj by line segments to pj-1, pj+1, and pj+k where   subscripts are to be interpreted modulo n. Thus, for example, if n = 18, then the point   p12 is joined to p11, p13, and p3 (since 12 + 9 + 3 (mod 18)) and the point p1 is joined   to p18 (since 1-1=18 (mod 18)), p2, and p10. You should convince yourself that this   construction for n even joins each point to exactly 3 others.   correct answer 18
<result>C</result>
Suppose that there are n points in the plane and that each of these
points is joined to exactly three of the others by straight line segments.
Then a possible value of n would be