jmc

The divisors of 42 are 1, 2, 3, 6, 7, 14, 21, and 42. They can be paired off into four pairs in such a way that the product of each pair is 42: $$\begin{array}{rl}\{1& ,42\},\\ \{2& ,21\},\\ \{3& ,14\},\text{and}\\ \{6& ,7\}.\end{array}$$Thus $A=42^4=(2\cdot 3\cdot 7{)}^4=2^4\cdot 3^4\cdot 7^4$ has $\boxed{3}$ prime divisors.