Let $A$ be the product of the divisors of $42$. How many distinct prime factors does $A$ have?
To find the product of the divisors of $42$, I need to list them first. The divisors of $42$ are $1, 2, 3, 6, 7, 14, 21,$ and $42$. To find the product of these numbers, I can pair them up as follows: $1 \cdot 42 = 42, 2 \cdot 21 = 42, 3 \cdot 14 = 42,$ and $6 \cdot 7 = 42$. So, $A = 42 \cdot 42 \cdot 42 \cdot 42 = 42^4$. So, $42^4 = (2 \cdot 3 \cdot 7)^4 = 2^4 \cdot 3^4 \cdot 7^4$. This means that $A$ has three distinct prime factors: $2, 3,$ and $7$. 