What is the maximum possible value of n?
$$30\cdot31\cdot32\ ...\ \cdot99=\frac{99!}{29!}$$
Highest power of 5 in 99! while neglecting the fractional part =>
$$\frac{99}{5}+\frac{99}{5^2}\ =\ 19.8+3.96$$
Highest power of 5 in 29! while neglecting the fractional part=>
$$\frac{29}{5}+\frac{29}{5^2}\ =\ 5.8+1.16\ =\ 6.96$$
Maximum possible value of n => Highest power of 5 in 99!/29! =
$$23.76-6.96=16.8\approx16$$

Since 16 and 18 are the answers which 16.8 falls in their range.
And whenever we are looking for the highest power of an integer in a factor, we must do away with the remainders.
So, approximating 16.8 without the remainder will give us 16.
Therefore, option D is the answer.

Thanks

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The multiples of 5 between 30 and 99, inclusive are 30, 35, ... , 95. There are (95-30)/5 + 1 = 14 such multiples. However, of these, two are also multiples of 25. These are 50 and 75. Each contributes an additional factor of 5 to the product (30)*(31)*(32)*...*(99).
It follows that the product is divisible by 5^14 *5*5 = 5^14*5^2 = 5^16.

Therefore, D is the answer.
Answer: D