2016 AMC 10A Problem 22

Write $n = 2^a 5^b 11^c \cdot k$ where $k$ has no factor of $2$, $5$ or $11$, and let the primes of $k$ have exponents $e_1, e_2, \dots$ in $n$. Then
$$110 n^3 = 2^{3a+1}\, 5^{3b+1}\, 11^{3c+1} \prod p_i^{3e_i},$$
so its divisor count is
$$(3a+2)(3b+2)(3c+2)\prod (3e_i + 1) = 110 = 2\cdot 5\cdot 11.$$

Each of the three factors $3a+2$, $3b+2$, $3c+2$ is at least $2$, so their product is at least $8$; the leftover $\prod(3e_i+1)$ must then be a divisor of $110$ that is at most $13$ and is $1$ more than a multiple of $3$. The candidates $1, 10$ fail unless the product is $1$ (with $10$ the remaining product would be $11$, too small to be a product of three numbers each at least $2$). So $k = 1$ and $\{3a+2, 3b+2, 3c+2\} = \{2, 5, 11\}$, i.e. $\{a, b, c\} = \{0, 1, 3\}$ in some order.

Thus $n = p\, q^3$ for two distinct primes $p, q \in \{2, 5, 11\}$, and neither is $3$. Then
$$81 n^4 = 3^4\, p^4\, q^{12},$$
which has $(4+1)(4+1)(12+1) = 5\cdot 5\cdot 13 = 325$ divisors, regardless of which primes play the roles of $p$ and $q$.

The answer is $\boxed{\textbf{(D)}\ 325}$.

The divisor-count function depends only on exponents, and multiplying by $110$ shifts three exponents by $1$. Cubing $n$ makes every exponent a multiple of $3$, so the factors of the divisor count fall into residue classes mod $3$ that distinguish "primes in $110$" from "other primes." Matching those classes against the factorization of $110$ pins down $n$'s exponent multiset without determining $n$ itself, and the final answer is symmetric in the choice, exactly what a well-posed problem needs.

• Assuming n must be a specific number (like n = 1 or n = 2 * 5^3) and not realizing the exponents can be permuted, or forgetting to include the 3^4 factor of 81 in the final count.
• Setting the divisor count of $110n^3$ equal to $110$ by using exponents $a, b, c$ directly rather than $3a+1$, $3b+1$, $3c+1$.