The number of the fifth-grade student is $$n$$.  $$N=[(n+2)\times(n+1)-20]\div n= (n^{2}+3n-18)\div n = n+3-\frac{18}{n}$$  $$n=18$$  $$N=18+3-1=20$$ 
Two sixth-grade students play chess with at least $$10$$ fifth-grade students, and each of the two men plays exactly one game against each other. There are three results: $$2$$ points for winning, $$1$$ point for drawing, and $$0$$ point for losing. After the competition, it is known that the sum of the two sixth grade students is $$20$$ points, and each fifth grader has scored $$N$$ points. $$N$$ is~\uline{~~~~~~~~~~}~. 
a: $$12$$ 
b: $$14$$ 
c: $$16$$ 
d: $$18$$ 
e: $$20$$ 