The sum of a two digit number and the number obtained by reversing the order of the digits is 99. Find the number, if the digits differ by 3.

Let digit at ten's place be x and unit's place be y.

Number = 10(x) + y = 10x + y

On reversing the digits,

Reversed number = 10(y) + x = 10y + x

Given,

Sum of a two digit number and the number obtained by reversing the order of the digits is 99.

∴ (10x + y) + (10y + x) = 99

⇒ 11x + 11y = 99

⇒ 11(x + y) = 99

⇒ x + y = 9

⇒ x = 9 - y …………(1)

Given,

Digits differ by 3.

∴ x - y = 3 or y - x = 3

Considering x - y = 3

⇒ x = 3 + y ……….(2)

From (1) and (2), we get :

⇒ 9 - y = 3 + y

⇒ 9 - 3 = 2y

⇒ 2y = 6

⇒ y = 3.

Substituting value of y = 3 in equation (1), we get :

⇒ x = 9 - 3 = 6.

Number = 10x + y = 10(6) + 3 = 60 + 3 = 63.

Considering y - x = 3

⇒ x = y - 3 ……….(3)

From (1) and (3), we get :

⇒ 9 - y = y - 3

⇒ 9 + 3 = 2y

⇒ 2y = 12

⇒ y = 6.

Substituting value of y = 6 in equation (1), we get :

⇒ x = 9 - 6 = 3.

Number = 10x + y = 10(3) + 6 = 30 + 6 = 36.

Hence, number = 36 or 63.