plz help on this num theory question

$\text{listing the divisors of 90}\\ 1,2,3,5,6,9,10,15,18,30,45,90\\ \text{We can collect pairs of divisors that multiply to 90}\\ (1,90),(2,45),(3,30),(5,18),(6,15),(9,10)\\ \text{We see then that the product is }90^6$

$\text{In general the product of the factors of }n \text{ will be }n^{\frac 1 2\text{# of factors of n}}\\ \text{You can determine the # of factors of }n \text{ by taking the product of each of the}\\ \text{exponents of the prime factors plus 1. For example }\\ 90 = 2^1\cdot 3^2 \cdot 5^1\\ \text{It has }(1+1)(2+1)(1+1) = 12 \text{ factors which multiply to }90^{12/2} = 90^6$

Factors of 90 =

 1 2 3 5  6 9 10 15 18 30  45  90

We  can write

(1 * 90) * (2* 45 )  *  (3 * 30)  * (5 * 18)   * (6 * 15 )  * ( 9 *10)  =

   90    *     90        *      90      *     90      *     90        *   90     =

90^6

k  =  6

We know that if $x$ is a factor of $n$, then $\frac{n}{x}$ is also a factor of $n$. Because $90$ is not a perfect square, we can pair up the $x$'s and the $\frac{n}{x}$'s and multiply them to get $n$. Because this is true for every number, we can count the number of factors of $90 = 2 \cdot 3^2 \cdot 5$. We can choose 2 ways for the exponent of 2, 3 for the exponent of 3, and 2 for the exponent of 5. So, there are $2 \cdot 3 \cdot 2 = 12$ factors. This means there are $\frac{12}{2}=6$ paris that multiply to $90$, and their products together is $90^6$, so the answer is $\boxed{k=6}$.

Hoping this helped, 

asdf334