Digit Product of 96

We are interested in finding $x,y,z \in \mathbb{N}$ such that $xyz = 96.$ Knowing that $96 = 2^{5}3^{1}$ , we have $6$ positive integer divisor pairs:

$(1,96); (2,48); (3,32); (4,24); (6,16); (8,12)$

We determine that $12 = 2\cdot 6 = 3 \cdot 4$ , which gives the triplet sets $(2,6,8)$ and $(3,4,8)$ (each of which produces $6$ unique three-digit numbers). Also, we determine that $16 = 2 \cdot 8 = 4 \cdot 4$ for the triplet sets $(6,2,8)$ and $6,4,4)$ (the former being a duplicate). Next, we have $24 = 3 \cdot 8 = 4 \cdot 6$ , or the triplets $(4,3,8); (4,4,6)$ (both are duplicates). There's also $32 = 4 \cdot 8$ which gives $(3,4,8)$ (another duplicate). Finally, there's $48 = 6 \cdot 8$ which yields $(2,6,8)$ (yet another duplicate).

Altogether, we have the critical triplets:

$(2,6,8) \Rightarrow 268,286,628,682,826,862$ ( $6$ total);

$(3,4,8) \Rightarrow 348, 384,438,483,834,843$ ( $6$ total);

$(4,4,6) \Rightarrow 446,464,644$ ( $3$ total)

or $\boxed{15}$ such numbers.