【Question 12】  
The 200-digit number $M$ consists of 200 ones, $M \times 2013$, the sum of the digits of the product is $\qquad$.
【Analysis and Solution】
$$
M \times 2013=M \times 2000+M \times 10+M \times 3=\underbrace{11 \cdots 1}_{200 \uparrow 1} \times 2000+\underbrace{11 \cdots 1}_{200 \uparrow 1} \times 10+\underbrace{11 \cdots 1}_{200 \uparrow 1} \times 3=\underbrace{22 \cdots 2}_{200 \uparrow 2} 000+\underbrace{11 \cdots 10}_{200 \uparrow 1}+\underbrace{33 \cdots 3}_{200 \uparrow 3} ;
$$
The sum of the digits of $\underbrace{22 \cdots 2}_{200 \uparrow 2} 000$ is $2 \times 200+0 \times 3=400$;
The sum of the digits of $\underbrace{11 \cdots 10}_{200 \uparrow 1}$ is $1 \times 200+0=200$;
The sum of the digits of $\underbrace{33 \cdots 3}_{200 \uparrow 3}$ is $3 \times 200=600$;

And $\underbrace{22 \cdots 2}_{200 \uparrow 2} 000+\underbrace{11 \cdots 10}_{200 \uparrow 1}+\underbrace{33 \cdots 3}_{200 \uparrow 3}$ does not produce a carry when added;
The sum of the digits of $\underbrace{22 \cdots 2}_{200 \uparrow 2} 000+\underbrace{11 \cdots 10}_{200 \uparrow 1}+\underbrace{33 \cdots 3}_{200 \uparrow 3}$ is $400+200+600=1200$;
The sum of the digits of the product $M \times 2013$ is 1200.