What's the number of SMS 679-3?

Is there anything wrong with this topic? It should be untenable, because only 1 to 9 does not make the equation hold.

First of all, we can turn this problem into an addition problem, that is, hundreds of 18 plus hundreds of 4 equals 610, which is numerically expressed as:

XX8+X4X=6XX (writing vertically is more intuitive)

Namely _ _ 8 ①

+ _ 4 _ ②

= 6 _ _ ③

So the remaining numbers are 1, 2, 3, 5, 7 and 9 respectively.

Because 9 is the largest, we should consider units and ten digits at first.

A. If the digits of ② are filled with 9, then the digits of ③ are 7, plus 1, then among the remaining 1, 2, 3, 5.

1+4+1= 6,2+4+1= 7,3+4+1= 8,5+4+1= 0, all of which do not meet the requirements of the topic, so it is assumed that A is not valid.

B Assuming that the unit of ② is 7, then the unit of ③ is 5, and you should also enter 1, then the remaining 1, 2, 3, 9.

1+4+1= 6,2+4+1= 7,3+4+1= 8,9+4+1=14, one digit is 4.

C Assuming that the unit of ② is filled with 5, then whether the unit of ③ is 3 or 1, then the remaining 1, 2, 7, 9.

1+4+1= 6,2+4+1= 7,7+4+1=12, take 2,9+4+1=/kloc.

In this way, 5, 3, 2 and 7 are used, but the remaining 1 and 9 can't add up to 6, so it is invalid to assume C.

D Assuming that the unit of ② is filled with 3, then the unit of ③ is 1, or it is one, then the remaining 2, 5, 7 and 9.

2+4+1 = 7,5+4+1=10, take 0,7+4+1=12, take 2, enter1,9+.

Only when 2 or 7 is used, then when it is 2, the ten digits in ③ are 7, and the remaining 5 and 9 cannot be added to 6.

Similarly, when using 7, the ten digits in ③ are 2. If you enter one digit, the remaining 5 and 9 cannot be added to 6, so it is assumed that D is invalid.

E If the digits of ② are filled with 2, 8+2= 10 will appear, and 0 will be taken, while the topic only requires 1 to 9, so it is assumed that E is not valid.

F If the unit of ② is filled with 1, then the unit of ③ is 9, so among the remaining 2, 3, 5 and 7.

2+4=6, 3+4=7, 5+4=9, 7+4= 1 1, take 1 and enter one.

We can only take 3+4=7, so we use 1, 9, 3, 7, but the remaining 2 and 5 can't add up to 6, so we assume that F is invalid.

To sum up, all the assumptions are invalid, that is, there is no solution according to the requirements of this question.