【Analysis】Since $2+0+1+1=4$ and $0+1+2+3+4+6+7+8+9=40, 4 \equiv 40(\bmod 9)$, the nine different Chinese characters represent the digits: $0,1,2,3,4,6,7,8,9$. It is easy to see: $40-4=36, 36 \div 9=4$ (times), indicating that the equation involves four carries. 
$$
“ 4=2+2=1+1+2=1+2+1 \text { ” }
$$

Obviously: (1) Hua $=1, “ 4=2+2$ ” has no solution
(2) Hua $=1, “ 4=1+1+2$ ” has a solution
According to this, we can analyze and solve the problem.

【Solution】Solution: Since $2+0+1+1=4$ and $0+1+2+3+4+6+7+8+9=40, 4 \equiv 40(\bmod 9)$, the nine different Chinese characters represent the digits: $0,1,2,3,4,6,7,8,9$.

It is easy to see: $40-4=36, 36 \div 9=4$ (times), indicating that the equation involves four carries.
$$
“ 4=2+2=1+1+2=1+2+1 \text { ” }
$$

Obviously: (1) Hua $=1, “ 4=2+2$ ” has no solution
(2) Hua $=1, “ 4=1+1+2$ ” has a solution

A: $28+937+1046=2011$, can form 36 equations $(6 \times 6 \times 1=36 ）$
B: $69+738+1204=2011$, can form 48 equations $(6 \times 4 \times 2=48 ）$
C: $79+628+1304=2011$, can form 48 equations $(6 \times 4 \times 2=48 ）$
(3) Hua $=1, “ 4=1+2+1$ ” has a solution

A: $46+872+1093=2011$, can form 36 equations $(6 \times 6 \times 1=36)$
B: $98+673+1240=2011$, can form 72 equations $(6 \times 6 \times 2=72)$
C: $97+684+1230=2011$, can form 72 equations $(6 \times 6 \times 2=72)$
Total: $72 \times 3+96=216+96=312$ (equations).
Answer: There are 312 in total.
13. (15 points) In the addition problem shown below, different Chinese characters represent different digits. How many different equations satisfy the requirement?



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