6. The product of all of the positive integer divisors of $6^{16}$ equals $6^{k}$ for some integer $k$. Determine the value of $k$.
6. Solution 1
We rewrite $6^{16}$ as $(2 \cdot 3)^{16}$, which equals $2^{16} 3^{16}$.
If $d$ is a positive integer divisor of $6^{16}$, it cannot have more than 16 factors of 2 or of 3 , and cannot include any other prime factors. Therefore, if $d$ is a positive integer divisor of $6^{16}$, then $d=2^{m} 3^{n}$ for some integers $m$ and $n$ with $0 \leq m \leq 16$ and $0 \leq n \leq 16$.
Since there are 17 possible values for each of $m$ and $n$, then there are $17 \times 17=289$ possible positive integer divisors $d$.
Since $6^{16}$ is a perfect square $\left(6^{8} \times 6^{8}\right)$, then every positive integer divisor other than its square
root $\left(6^{8}\right)$ can be paired with a different positive integer such that the product of the pair is $6^{16}$. (For example, $2^{4} 3^{13}$ can be paired with $2^{12} 3^{3}$ since $2^{4} 3^{13} \times 2^{12} 3^{3}=2^{4+12} 3^{13+3}=2^{16} 3^{16}$.)
Since there are 289 positive integer divisors in total, then there are $\frac{1}{2}(289-1)=144$ of these divisor pairs whose product is $6^{16}$.
To multiply all of the positive integer divisors together, we can multiply the product of the 144 pairs with the remaining divisor $\left(6^{8}\right)$.
Therefore, the product of all of the positive integer divisors is $\left(6^{16}\right)^{144} \cdot 6^{8}=6^{16(144)+8}=6^{2312}$. Thus, $k=2312$.

Solution 2
We rewrite $6^{16}$ as $(2 \cdot 3)^{16}$, which equals $2^{16} 3^{16}$.
If $d$ is a positive integer divisor of $6^{16}$, it cannot have more than 16 factors of 2 or of 3 , and cannot include any other prime factors. Therefore, if $d$ is a positive integer divisor of $6^{16}$, then $d=2^{m} 3^{n}$ for some integers $m$ and $n$ with $0 \leq m \leq 16$ and $0 \leq n \leq 16$.
Consider the 17 divisors of the form $2^{0} 3^{n}$ for $n=0$ to $n=16$.
These divisors are $3^{0}, 3^{1}, 3^{2}, \ldots, 3^{15}, 3^{16}$. Their product is
$$
3^{0} \cdot 3^{1} \cdot 3^{2} \cdots \cdot 3^{15} \cdot 3^{16}=3^{0+1+2+\cdots+15+16}
$$

By the Useful Fact given, $0+1+2+\cdots+15+16=1+2+\cdots+15+16=\frac{1}{2}(16)(17)=136$, and so the product of these divisors is $3^{136}$.
Next, consider the 17 divisors of the form $2^{1} 3^{n}$ for $n=0$ to $n=16$. These divisors are $2^{1} 3^{0}, 2^{1} 3^{1}, 2^{1} 3^{2}, \ldots, 2^{1} 3^{15}, 2^{1} 3^{16}$. Their product is
$$
2^{1} 3^{0} \cdot 2^{1} 3^{1} \cdot 2^{1} 3^{2} \cdots \cdot 2^{1} 3^{15} \cdot 2^{1} 3^{16}=2^{17} 3^{0+1+2+\cdots+15+16}=2^{17} 3^{136}
$$

In general, consider the 17 divisors of the form $2^{m} 3^{n}$, for a fixed value of $m$ and for $n=0$ to $n=16$. These divisors are $2^{m} 3^{0}, 2^{m} 3^{1}, 2^{m} 3^{2}, \ldots, 2^{m} 3^{15}, 2^{m} 3^{16}$. Their product is
$$
2^{m} 3^{0} \cdot 2^{m} 3^{1} \cdot 2^{m} 3^{2} \cdots \cdots 2^{m} 3^{15} \cdot 2^{m} 3^{16}=2^{17 m} 3^{0+1+2+\cdots+15+16}=2^{17 m} 3^{136}
$$

To obtain the overall product of all of the positive integer divisors, we can multiply together the products of these sets of 17 divisors.
This product is
$$
\begin{aligned}
2^{0} 3^{136} & \cdot 2^{17(1)} 3^{136} \cdot 2^{17(2)} 3^{136} \cdots \cdots 2^{17(15)} 3^{136} \cdot 2^{17(16)} 3^{136} \\
& =2^{17(0+1+2+\cdots+15+16)}\left(3^{136}\right)^{17} \\
& =2^{17(136)} 3^{17(136)} \\
& =2^{2312} 3^{2312} \\
& =6^{2312}
\end{aligned}
$$

Therefore, the desired product is $6^{2312}$, and so $k=2312$.
ANSWER: 2312