How many digits are there in the number $2^3 \cdot 3^1 \cdot 4^3 \cdot 5^8$ ?
We want to pair together the twos and the fives to make tens: \begin{align*}
2^3 \cdot 3^1 \cdot 4^3 \cdot 5^8 &=2^3 \cdot 3 \cdot 2^6 \cdot 5^8 \\
&=2^9 \cdot 3 \cdot 5^8 \\
&=10^8 \cdot 2 \cdot 3 \\
&=6 \cdot 10^8
\end{align*}Therefore, we have $6$ with $8$ zeros at behind it, giving us $1+8=\boxed{9}$ digits.