The sum of a two digit number and the number obtained by interchanging the digits of the number is 121. If the digits of the number differ by 3, find the number.

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

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

On reversing digits, the number becomes = 10(y) + x = 10y + x.

Given,

The sum of a two digit number and the number obtained by interchanging the digits of the number is 121.

∴ 10x + y + 10y + x = 121

⇒ 11x + 11y = 121

⇒ 11(x + y) = 121

⇒ x + y = 11 ……….(1)

Let digits of the number differ by 3.

⇒ x - y = 3 ……..(2)

or

⇒ y - x = 3……….(3)

Considering x - y = 3,

Adding equation (1) and (2), we get :

⇒ x + y + x - y = 11 + 3

⇒ 2x = 14

⇒ x = 7.

Substituting value of x in equation (1), we get :

⇒ 7 + y = 11

⇒ y = 11 - 7

⇒ y = 4.

Number = 10x + y = 10(7) + 4 = 70 + 4 = 74.

Considering y - x = 3,

Adding equation (1) and (3), we get :

⇒ x + y + y - x = 11 + 3

⇒ 2y = 14

⇒ y = $\dfrac{14}{2}$

⇒ y = 7.

Substituting value of x in equation (1), we get :

⇒ x + 7 = 11

⇒ x = 11 - 7

⇒ x = 4.

Number = 10x + y = 10(4) + 7 = 40 + 7 = 47.

Hence, number = 47 or 74.