2016 AMC 10A Problem 22

The prime factorization is $110=2\cdot5\cdot11$. If $110n^3$ has $110=2\cdot5\cdot11$ divisors, then it has exactly three prime factors, with exponents $1,4,10$ in some order. For each of the primes $2,5,11$, its exponent in $110n^3$ is $1$ more than a multiple of $3$. The exponents $1,4,10$ all have this form, so the corresponding exponents in $n$ are $0,1,3$ in some order. In $81n^4=3^4n^4$, the exponents from $n^4$ are therefore $0,4,12$, in some order, along with the exponent $4$ on prime $3$. Hence the divisor count is $$\begin{aligned} &(0+1)\cdot(4+1)\cdot(12+1) \\ &\quad {}\cdot(4+1)=325. \end{aligned}$$ Thus, the correct answer is D.