Problem 1705

For a positive integer $k > 1$, let $f(k)$ represent the number of ways to factor $k$ into a product of positive integers greater than 1. For example, $f(12) = 4$ because 12 can be factored in these 4 ways: $12$, $2 \cdot 6$, $3 \cdot 4$, and $2 \cdot 2 \cdot 3$.

We aim to prove that if $n$ is a positive integer greater than 1 and $p$ is a prime factor of $n$, then $f(n) \leq \frac{n}{p}$.

We proceed by using strong induction. The base case is clear.

Let $p$ be the largest prime divisor of $n$. We need to show that $f(n) \leq \frac{n}{p}$.

Consider $n = \prod x_j$ where each $x_j > 1$. Suppose one of the factors $x_i = p \cdot d_1$. Then $d_1$ must divide $\frac{n}{p}$, implying that:
$$f(n) \leq \sum_{d_1 \mid \frac{n}{p}} f\left(\frac{n}{p d_1}\right).$$

By the inductive hypothesis, for any $k < n$, we have $f(k) \leq \frac{k}{Q(k)} \leq \phi(k)$, where $Q(k)$ is the largest prime factor of $k$ and $\phi(k)$ is the Euler's totient function.

Thus, we can write:
$$f(n) \leq \sum_{d_1 \mid \frac{n}{p}} f\left(\frac{n}{p d_1}\right) \leq \sum_{d_1 \mid \frac{n}{p}} \phi\left(\frac{n}{p d_1}\right) = \frac{n}{p}.$$

Therefore, we have shown that $f(n) \leq \frac{n}{p}$ as required.

The answer is: \boxed{\frac{n}{p}}.