The sum of the digits of a 2-digit number is 11. If the number obtained by reversing the digits is 9 less than the original number, find the number.

Let ab be the 2-digit number.

Given that a + b = 11 and ab – ba = 9

In general, a 2-digit number ab made of digits a and b is equal in value to 10 × a + b = 10 a + b

Similarly, we have ba = 10 × b + a = 10b + b

From the problem statement, we know ab – ba = 9

(10a + b) – (10b + a) = 9

9a – 9b = 9

⇒ a – b = 1

So, we have a + b = 11 and a – b = 1. Adding these,

(a + b) + (a – b) = 11 + 1

⇒ 2a = 12

∴ a = 6

Using a – b = 1, we have b = a – 1

∴ b = 6 – 1 = 5

So, ab = 65

Therefore, the number is 65.