ibm - the sum of the digits of a two digit number is equal to 9 and the number obtained by interchanging the digits of that number is 63 less than the number then the tens place of the two digit number is - skillgun

Let the ten’s place of the two digit number be x.
Since the sum of the digit is 9, the unit's place will be 9-x.
Then the number is represented by = 10x +(9-x) = 9x + 9
For the number with digits reversed, tens's place will be 9 - x and unit's place will be 9.
Therefore, the reversed digits number can be represented as 10(9-x)+x = 90 - 9x
Given that the new number is 63 less than the original number.
Therefore, New Number = Original Number - 63 i.e 90-9x = 9x + 9 - 63
18x = 144
x = 144/18
x = 8.
Then the number is 9(8) + 9 = 81.
Hence the required digit is 8.