Let the digits at tens place and ones place be x and 9 − x respectively.
Therefore, original number = 10x + (9 − x) = 9x + 9
On interchanging the digits, the digits at ones place and tens place will be x and 9 − x respectively.
Therefore, new number after interchanging the digits = 10(9 - x) + x
= 90 - 10x + x
= 90 - 9x
90 - 9x = 9x + 9 + 27
90 - 9 = 9x + 36
Transposing 9x to R.H.S and 36 to L.H.S, we obtain
90 - 36 = 18x
54 = 18x
Dividing both sides by 18, we obtain
3 = x and 9 - x = 6
Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.
Therefore, the two-digit number is 9x + 9 = 9 × 3 + 9 = 36
