For example, 1 to 6 should be filled in the six spaces of three sides of a triangle, so that the sum of each side is equal. The solution is 1+2+3+4+5+6=2 1, and 2 1 is divisible by 3, so the sum of the three vertices is also a multiple of 3. From 6 numbers from 1 to 6, choose 3 numbers to make the sum a multiple of 3.
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Because the sum of each row is required to be 12, and the number given is not 1, and the position of 2 in the question has been fixed, it is necessary to ensure that the sum of the two numbers is less than 1 1. Of the three numbers given, 6+7= 13, 5+7= 12, 5+6= 1 1, 4+7 =1.
After exclusion, there are still combinations of 3+4 = 7, 3+5 = 8, 3+6 = 9, 3+7 = 10, 4+5+9 and 4+6= 10. According to the known position of 2, we can also determine 3, 7, 4 and 6.