The largest $$n\epsilon N$$, for which $$7^{n}$$ divides 101!, is :

To calculate the largest power of 7 which divides 101!, we find the largest integer by dividing 101 by the increasing the power of the prime number (7) by one in subsequent step till we get 0. Then we will add all the largest integers to get the largest power of 7.  

Largest power of 7 which divides 101! = $$\left[\dfrac{101}{7}\right]+\left[\dfrac{101}{7^2}\right]+\left[\dfrac{101}{7^3}\right]+\dots+\infty$$

$$\left[\dfrac{101}{7}\right]=14$$ as the largest integer which satisfies it is 14 (greatest integer function). 

$$\left[\dfrac{101}{7^2}\right]=\left[\ \dfrac{101}{49}\right]=2$$

$$\left[\dfrac{101}{7^3}\right]=\left[\ \dfrac{101}{343}\right]=0$$

Now, the value of all the greatest integer funtion following this will be 0. 

So, the largest power of 7 which divides 101! is $$(14+2+0)=16$$.

$$\therefore\ $$ The required answer is D.