12 Positive integer $n \leqslant 500$, has the following property: For any element $m$ chosen from the set $\{1,2, \cdots, 500\}$, the probability that $m$ divides $n$ is $\frac{1}{100}$, then the maximum value of $n$ is $\qquad$ .
12 From the problem, we know that $n$ has exactly 5 positive divisors. Let the prime factorization of $n$ be $n=p_{1}^{a_{1}} \cdots p_{k_{k}}^{q_{k}}$, then the number of positive divisors of $n$ is $\left(\alpha_{1}+1\right) \cdots\left(\alpha_{k}+1\right)$, so
$$
\left(\alpha_{1}+1\right) \cdots\left(\alpha_{k}+1\right)=5,
$$

thus $n$ must be of the form $p^{4}$, where $p$ is a prime. Since $3^{4}=81,5^{4}=625>500$, the maximum value of $\pi^{2}$ is 81.