5. Find the sum of the digits of the number $\underbrace{44 \ldots 4}_{2012 \text { times }} \cdot \underbrace{99 \ldots 9}_{2012 \text { times }}$
Answer: 18108.

Solution: Note that $\underbrace{4 \ldots 4}_{2012} \cdot \underbrace{9 \ldots 9}_{2012}=\underbrace{4 \ldots 4}_{2012} \underbrace{0 \ldots 0}_{2012}-\underbrace{4 \ldots 4}_{2012}=\underbrace{4 \ldots 4}_{2011} 3 \underbrace{5 \ldots 5}_{2011} 6$. The sum of the digits is $4 \cdot 2011+3+5 \cdot 2011+6=18108$.