All those factors!!!

Let's generalize at first. Suppose we need to find the product of all positive factors of $n$ with $k$ factors. We'll use a simple fact: if $d\mid n$ , then $n /d \mid n$ and $d\cdot n/ d = n$ . So we can pair them up. In case of a non-perfect square, the number of factors of a natural is even. So we have $k /2$ such pairs. And hence, the product of all factors of $n$ is given by:

$n^{k/2} ~~~~~~\left[n=\prod_{a_i\in\mathbb{N_0}}^{p_i \in \mathbb{P}}{p_i}^{a_i}, ~~~ k=\prod (a_i + 1) \right]$

In case of a perfect square the number of factors is odd. So we have $(k-1)/2$ pairs. Hence, the product of all factors of $n$ is given by:

$n^{(k-1)/2}\cdot \sqrt{n}=n^{(k-1)/2}\cdot n^{1/2}=n^{k/2}~~~~~~~~\left[n=\prod_{a_i\in\mathbb{N_0}}^{p_i \in \mathbb{P}}{p_i}^{a_i}, ~~~ k=\prod (a_i + 1) \right]$

Therefore, in both case, the product is $n^{k/2}$ . Let's come back to earth, we are given $144=2^4\cdot 3^2$ which has $k=(4+1)(2+1)=15$ factors. Our desired answer is:

$144^{15/2}=12^{15}=2^{30}\cdot 3^{15}\equiv 824 \cdot 907\equiv \fbox{368}\pmod{1000}$