For a two digit number, the sum of the number on the tenth digit and the number on the single digit is 11. If the number on the tenth digit and the number on the single digit are exchanged, the new number will be 63 larger than the original number, and the original two digits will be obtained

For a two digit number, the sum of the number on the tenth digit and the number on the single digit is 11. If the number on the tenth digit and the number on the single digit are exchanged, the new number will be 63 larger than the original number, and the original two digits will be obtained

Let the original digit be x, then the number on the ten digit is 11-x. according to the meaning of the question, 10 × (11-x) + x = 63 + 10x + (11-x), the solution is: x = 2, the number on the original ten digit is 9, that is, the original two digit 29

Two digit, the sum of the number of ten digit and the number of individual digit is 11. If the number of ten digit is transposed with the number of individual digit, the new number is 63 larger than the original number, and the original number is calculated

Let the original tens of digits be x, then the ones are 11-x
10（11-x）+x-（10x+11-x）=63
The solution is x = 2
A: the original number is 29

The sum of the numbers in a three digit number is 11. If you transpose the number in a hundred with the number in a single digit, the number you get is 693 larger than the original number
The one digit is x, the ten digit is y, and the hundred digit is Z

128 or 209
99x-99z=693;
So x-z = 7; Z is not zero, so x = 8 or 9
When x = 8, it is 128
When x = 9, it is 209