Let the one’s digit be ‘a’ and ten’s digit be ‘b’

Given, sum of two-digit number and the number formed by reversing the order of digits is 66.

⇒ 10a + b + 10b + a = 66

⇒ a + b = 6

Also, digits differ by 2.

⇒ a – b = 2 or b – a = 2

Adding both equation

2a = 8 or 2b = 8

⇒ a = 4 and b = 2 or a = 2 and b = 4

Number can be 24 or 42