2016 AMC 12A Problem 18

$110=2\cdot5\cdot11,$ so $110n^3$ has exactly three prime factors

The exponents plus one multiply to $110=2\cdot5\cdot11;$ deduce $n=p\cdot q^3$

The number $110n^3$ is divisible by the three distinct primes $2,5,11.$ If its prime exponents are $r_1,r_2,\ldots,$ then $(r_1+1)(r_2+1)\cdots=110,$ where $110=2\cdot5\cdot11.$ Because there are already at least three factors greater than $1,$ there are exactly three, so no other prime divides $n.$

Each exponent in $110n^3$ is $1\pmod3,$ so each divisor-count factor $r_i+1$ is $2\pmod3.$ The factors $2,5,11$ all have that form, so the exponents are $1,4,10$ in some order. After subtracting the exponent $1$ from $110$ and dividing by $3,$ the exponents in $n$ are $0,1,3$ in some order. Thus $n=pq^3$ for two distinct primes $p,q$ chosen from $2,5,11.$

Consequently $n^4$ has exponents $4$ and $12,$ while $81=3^4$ introduces a third prime because $n$ is not divisible by $3.$ Hence the number of divisors of $81n^4$ is $$\begin{gathered} (4+1)(4+1)(12+1)\\ =5\cdot5\cdot13\\ =325. \end{gathered}$$

Thus, the correct answer is D.