Let the digit in units place be \\\\(x\\\\). Then, digit in ten's place \\\\(= 3x\\\\)So, original number \\\\(= 10(3x) + x = 31x\\\\)On interchanging the digits, new number \\\\(= 10x + 3x = 13x\\\\)According to the given condition,\\\\( {31x + 13x = 88} \\\\)\\\\(\\Rightarrow 44x=88\\\\ \\Rightarrow x=2\\\\)So, the original number is \\\\(31x\\\\) or \\\\(31 \ imes 2 = 62\\\\).On the other hand, if we consider the digit in ten's place as \\\\(x\\\\), then the digit in the unit's place will be \\\\(3x\\\\).So, original number \\\\(= 10(x) + 3x = 13x\\\\)On interchanging the digits, new number \\\\(= 10(3x) + x = 31x\\\\)According to the given condition,\\\\( {13x + 31x = 88} \\\\)\\\\(\\Rightarrow 44x=88\\\\ \\Rightarrow x=2\\\\)So, the original number is \\\\(13x\\\\) or \\\\(13 \ imes 2 = 26\\\\).Here, both answers are correct, as \\\\(26 + 62 = 88\\\\).