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Mathematical addition problem
These eight numbers make up four numbers. First of all, they can't round up three digits, otherwise they must be greater than 160, so they can only round up four two digits.

So four of these eight numbers are ten digits and four are single digits.

The sum of these eight numbers is 2+3+...+8+9 = 44.

Let the sum of digits of ten digits be x and x be an integer, then the sum of digits of one digit is (44-x).

So x *10+(44-x) =1609x =116.

x = 1 16/9 = 12.88

Contradicting that x is an integer.

So there is no solution.

I hope this helps.