13. (15 points) In the addition problem shown below, different Chinese characters represent different digits. How many different equations satisfy the requirement?



The translation is provided while preserving the original text's line breaks and format.
【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.