The maximum value of the natural number n for which $$21^{n}$$ divides 50! is

$$21^n=3^n\times\ 7^n$$

Now for the maximum value of $$n$$ so that $$21^n$$ divides 50!, we can say we need to maximum value of n such that both $$3^n$$ and $$7^n$$ will divide 50!

Now, the exponent of prime p in n! is given by the formula,

$$E\left(p\right)=\left[\dfrac{n}{p}\right]+\left[\dfrac{n}{p^2}\right]+......$$

So, the exponent of 7 in 50! = $$=\left[\dfrac{50}{7}\right]+\left[\dfrac{50}{7\times\ 7}\right]+......$$ = $$7+1=8$$

Clearly 3 will have an exponent greater than 8 as 3 is less than 7

So, the maximum value of the natural number $$n$$ for which $$21^{n}$$ divides 50! is = 8