14. (15 points) In the following addition problem, different Chinese characters can represent the same digit. How many different equations satisfy the requirement?



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【Analysis】According to the calculation method of integer addition.
【Solution】From the vertical form, we can get: “Hua” $=1$;
Since in the addition vertical form, different Chinese characters can represent the same number;
Therefore, the sum of “Yue” + “Ri” + “Sai” in the units place is $21$, $11$ or 1;
The sum of “Yue” + “Ri” + “Sai” in the units place is 21, carry 2 to the tens place;
In the tens place, $4+6+$ “Jue” +2 ends with 1, by $4+6+9+2=21$, we get “Jue” $=9$, carry 2 to the hundreds place; in the hundreds place, $1+$ “Bei” +2 ends with 0, by $1+7+2=10$, we get “Bei” $=7$, carry 1 to the thousands place;

In the thousands place, $1+1$ is exactly 2;
From the above, as long as the sum in the units place is 21, “Hua”, “Bei”, and “Jue” are fixed numbers;
Similarly, if the sum of “Yue” + “Ri” + “Sai” in the units place is 11, we get “Hua” $=1$, “Bei” $=9$, “Jue” $=0$, which are also fixed numbers;

If the sum of “Yue” + “Ri” + “Sai” in the units place is 1, we get “Hua” $=1$, “Bei” $=9$, “Jue” $=1$, which are also fixed numbers;

Therefore, “Yue”, “Ri”, and “Sai” determine different equations;
(1) “Yue” + “Ri” + “Sai” $=21$;
$7+7+7=21$, we get 1 type;
$6+7+8=21$, we get 6 types;
$6+6+9=21$, we get 3 types;
$5+8+8=21$, we get 3 types;
$5+7+9=21$, we get 6 types;
$4+8+9=21$, we get 6 types;
$3+9+9=21$, we get 3 types;
Thus, the sum of “Yue” + “Ri” + “Sai” being 21 can result in $1+6+3+3+6+6+3=28$ different equations;
(2) “Yue” + “Ri” + “Sai” $=11$;
$2+0+9=11$, we get 6 types;
$3+0+8=11$, we get 6 types;
$4+0+7=11$, we get 6 types;
$5+0+6=21$, we get 6 types;
$1+1+9=11$, we get 3 types;
$2+1+8=11$, we get 6 types;
$3+1+7=11$, we get 6 types;
$4+1+6=11$, we get 6 types;
$5+1+5=11$, we get 3 types;
$2+2+7=11$, we get 3 types;
$3+2+6=11$, we get 6 types;
$4+2+5=11$, we get 6 types;
$3+3+5=11$, we get 3 types;
$4+3+4=11$, we get 3 types;
Thus, the sum of “Yue” + “Ri” + “Sai” being 11 can result in $6+6+6+6+3+6+6+6+3+3+6+6+3+3=69$ different equations;
(3) “Yue” + “Ri” + “Sai” $=1$;
$0+0+1=1$, we get 3 types;
Thus, the sum of “Yue” + “Ri” + “Sai” being 1 can result in 3 different equations;
In summary, there are a total of $28+69+3=100$ different equations.