Harvard-MIT November Tournament

Solution:

Answer: $27$

Note that $2310 = 2 \times 3 \times 5 \times 7 \times 11$. In general, we see that if $n$ has $d(n)$ positive integer factors, then $p(n) = n^{\frac{d}{2}}$ since we can pair factors $(d, \frac{n}{d})$ which multiply to $n$. As a result, $p(n)$ is a square if and only if $n$ is a square or $d$ is a multiple of $4$.

Thus, because $2310$ is not divisible by the square of any prime, we claim that for integers $n$ dividing $2310$, $p(n)$ is even if and only if $n$ is not prime. Clearly, $p(n)$ is simply equal to $n$ when $n$ is prime, and $p(1) = 1$, so it suffices to check the case when $n$ is composite. Suppose that $n = p_1 p_2 \cdots p_k$, where $k > 1$ and $\{p_1, \ldots, p_k\}$ is some subset of $\{2, 3, 5, 7, 11\}$. Then, we see that $n$ has $2^k$ factors, and that $4 \mid 2^k$, so $p(n)$ is a square.

Since $2310$ has $2^5 = 32$ factors, five of which are prime, $27$ of them have $p(n)$ even.