Solution

a) From the factorization of $n$, we can determine the factorization of $n^{2}$:

$$
\begin{aligned}
n^{2} & =\left(2^{7} \cdot 3^{4}\right)^{2} \\
& =2^{14} \cdot 3^{8}
\end{aligned}
$$

Thus, the number $n^{2}$ has $(14+1)(8+1)=15 \cdot 9=135$ positive divisors.

b) Note that all positive divisors of $n^{2}$ can be organized into pairs of distinct numbers of the form $\left(d, \frac{n^{2}}{d}\right)$ whose product is $n^{2}$, with the exception of $n$. Moreover, since the numbers in each pair are distinct and their product is $n^{2}$, one of them is greater than $n$ and the other is less than $n$. For example, if we consider $n=6$, the divisors of $n^{2}=36$ different from 6 can be organized into the following pairs: $\{1,36\},\{2,18\},\{3,12\}$ and $\{4,9\}$. Thus, we can conclude that the number of divisors of $n^{2}$ greater than $n$ is equal to the number of pairs we can form. Therefore, there are $\frac{135-1}{2}=67$ divisors of $n^{2}$ greater than $n$.

c) Let's continue the reasoning from the previous item. Observe that in each pair we have $x \cdot y=$ $n^{2}$, where $x<n<y$. By examining the prime factors of $x$ and $y$, $y$ is a multiple of $n$ in the cases where $x$ is a divisor of $n$ less than $n$. Since $n=2^{7} \cdot 3^{4}$ has $(7+1)(4+1)=40$ positive divisors, there are 39 of them different from $2^{7} \cdot 3^{4}$. Therefore, excluding the pairs where $x$ is a divisor of $n$ less than $n$, we can conclude that there are $67-39=28$ divisors of $n^{2}$ greater than $n$ that are not multiples of $n$.
To determine the number of positive divisors of a number, it is enough to factor it as powers of distinct primes and multiply the successors of the exponents. For example, $2016=2^{5} \cdot 3^{2} \cdot 5^{1}$ has $(5+1)(2+1)(1+1)=36$ positive divisors. Consider the number $n=2^{7} \cdot 3^{4}$.

a) Determine the number of positive divisors of $n^{2}$.

b) How many divisors of $n^{2}$ are greater than $n$?

c) How many divisors of $n^{2}$ are greater than $n$ and are not multiples of $n$?