The largest n∈𝑁, for which 7^𝑛 divides 101 ! is

The correct option is 16.

To find the largest natural number $n$ such that $7n$ divides $\mathrm{101!}$, we need to find the exponent of the prime number 7 in the prime factorization of $\mathrm{101!}$ (101 factorial).

We can use Legendre's formula to calculate the highest power of a prime $p$ that divides $\mathrm{n!}$. The formula is given by:

$E\left(p\right)=\lfloor \frac{n}{p}\rfloor +\lfloor \frac{n}{p^2}\rfloor +\lfloor \frac{n}{p^3}\rfloor +\dots$

Here, $n=101$ and $p=7$. Let's compute each term step-by-step:

1. First power of 7:
$\lfloor \frac{101}{7}\rfloor =\lfloor 14.428\rfloor =14$

2. Second power of 7 ($7^2=49$):
$\lfloor \frac{101}{49}\rfloor =\lfloor 2.061\rfloor =2$

3. Third power of 7 ($7^3=343$):
$\lfloor \frac{101}{343}\rfloor =0$

Now, summing up these values to find the total exponent $n$:

$n=14+2+0=16$

Thus, the largest natural number $n$ for which $7^n$ divides $\mathrm{101!}$ is 16.