分析：利用$$\left\lfloor x \right\rfloor $$表示不超过$$x$$的最大整数的性质，求出$$\left\lfloor \frac{n}{2} \right\rfloor +\left\lfloor \frac{n}{3} \right\rfloor +\left\lfloor \frac{n}{6} \right\rfloor $$的范围．  $$\because \left\lfloor \frac{n}{2} \right\rfloor \leqslant \frac{n}{2}$$，$$\left\lfloor \frac{n}{3} \right\rfloor \leqslant \frac{n}{3}$$，$$\left\lfloor \frac{n}{6} \right\rfloor \leqslant \frac{n}{6}$$$$\therefore \left\lfloor \frac{n}{2} \right\rfloor +\left\lfloor \frac{n}{3} \right\rfloor +\left\lfloor \frac{n}{6} \right\rfloor \leqslant \frac{n}{2}+\frac{n}{3}+\frac{n}{6}=n$$．  又$$\because \left\lfloor \frac{n}{2} \right\rfloor +\left\lfloor \frac{n}{3} \right\rfloor +\left\lfloor \frac{n}{6} \right\rfloor =n$$  $$\therefore \left\lfloor \frac{n}{2} \right\rfloor =\frac{n}{2}$$，$$\left\lfloor \frac{n}{3} \right\rfloor =\frac{n}{3}$$，$$\left\lfloor \frac{n}{6} \right\rfloor =\frac{n}{6}$$  $$\therefore n$$的个数有$$\left\lfloor \frac{100}{6} \right\rfloor =16$$个，选D． 
正整数$$n$$小于$$100$$，并满足等式$$\left\lfloor \frac{n}{2} \right\rfloor +\left\lfloor \frac{n}{3} \right\rfloor +\left\lfloor \frac{n}{6} \right\rfloor =n$$，其中$$\left\lfloor x \right\rfloor $$表示不超过$$x$$的最大整数，这样的正整数$$n$$有(  )个.
a: $$2$$个 
b: $$3$$个 
c: $$12$$个 
d: $$16$$个 