Counting the special numbers

Since such number $N$ is a multiple of $6^{57}$ , let $N = 6^{67}n$ , where $n$ is a natural number. Since $N$ divides $6^{75}$ , we have $\dfrac {6^{75}}N = m,$ where $m$ is a natural number. Then we have:

$\frac {6^{75}}{6^{57}n} = m, \implies mn = 6^{18} = 2^{18}\cdot 3^{18}$

Then the number of $N$ 's is $(18+1)(18+1) = \boxed{361}$ .

The "special numbers" in the question all have the form $2^a 3^b$ where $57 \le a \le 75$ and $57 \le b \le 75$ .

There are $19$ possible values for $a$ , and $19$ for $b$ ; we can choose these independently, so in total there are $19\times 19=\boxed{361}$ special numbers.