Given that a four digit number plus the sum of its digits is equal to 2008, then how much is the sum of all such four digits

Given that a four digit number plus the sum of its digits is equal to 2008, then how much is the sum of all such four digits

Suppose four digits (ABCD) = 1000A + 100b + 10C + D, then 1000A + 100b + 10C + D + A + B + C + D = 2008
Obviously a = 1 or a = 2,
When a = 2, B = C = 0, we can get d = 2, that is, 2002
When a = 1, B = 9, if C ≤ 6, then x ≥ 32,
If C = 7 or C = 9, we can get: odd = even, rounding
When C = 8, the solution is x = 5, that is 1985
So the sum of all these four digits is 2002 + 1985 = 3987

Six times of a number plus eight equals eight times of it minus six______ ．

Let this number be X. according to the meaning of the question, we can get the equation: 6x + 8 = 8x-6, & nbsp; & nbsp; 2x = 14, & nbsp; & nbsp; & nbsp; X = 7

Five times a number plus 15 is eight times that number. Find the number

5x + 15 = 8x
15 = 3x
x=5