### Instruction: The product of the positive integer divisors of a positive integer $n$ is 729. Find $n$.### Response: To solve this problem, I will use the fact that the product of the positive integer divisors of a positive integer $n$ is equal to $n^{d(n)/2}$, where $d(n)$ is the number of divisors of $n$.
I will also use the fact that 729 is a power of 3, namely $3^6$.
So, I can write the equation $n^{d(n)/2} = 3^6$ and try to find a positive integer $n$ that satisfies it.
Since $n$ must be a divisor of $3^6$, it can only have prime factors of 3.
So, I can write $n = 3^k$ for some nonnegative integer $k$.
Then, the number of divisors of $n$ is $d(n) = k + 1$, since the divisors are $1, 3, 3^2, \dots, 3^k$.
Plugging this into the equation, I get $(3^k)^{(k+1)/2} = 3^6$.
Simplifying, I get $3^{k(k+1)/2} = 3^6$.
Taking the logarithm base 3 of both sides, I get $k(k+1)/2 = 6$.
Multiplying both sides by 2, I get $k(k+1) = 12$.
I can solve this quadratic equation by factoring: $k^2 + k - 12 = 0$.
The factors are $(k + 4)(k - 3) = 0$.
So, the possible values of $k$ are $k = -4$ and $k = 3$.
However, $k = -4$ does not make sense, since $n = 3^k$ would not be a positive integer.
So, the only valid value of $k$ is $k = 3$.
Therefore, $n = 3^3 = 27$.