What is the remainder when the product of the $5$ smallest prime numbers is divided by $42$?
To find the product of the $5$ smallest prime numbers, I need to list them in order: $2, 3, 5, 7, 11$. The product is $2 \times 3 \times 5 \times 7 \times 11 = 2310$. To find the remainder when dividing by $42$, I can use long division or a calculator. I'll use a calculator and write $2310 = 42 \times 55 + r$, where $r$ is the remainder. Using a calculator, I get $r = 0$. This means that $2310$ is a multiple of $42$ and leaves no remainder when divided by it. 