A number consist of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

#### Solution

Let the digits at units and tens place of the given number be *x* and *y* respectively. Thus, the number is `10 y + x`.

The sum of the digits of the number is 5. Thus, we have `x + y = 5`

After interchanging the digits, the number becomes `10 x + y`.

The number obtained by interchanging the digits is greater by 9 from the original number. Thus, we have

`10 x + y = 10 y + x + 9`

`⇒ 10 x + y - 10y - x =9`

` ⇒ 9x -9 y = 9 `

`⇒ 9 ( x - y)= 9`

` ⇒ x - y = 9/9`

` ⇒ x - y = 1`

Here *x* and *y* are unknowns. We have to solve the above equations for *x* and *y*.

Adding the two equations, we have

`( x + y) + ( x - y) = 5 + 1`

`⇒ x + y + x - y = 6`

` ⇒ 2x = 6`

` ⇒ x = 6/2`

` ⇒ x = 3`

Substituting the value of *x *in the first equation, we have

` 3 + y = 5`

`⇒ y = 5-3`

` ⇒ y = 2`

Hence, the number is ` 10 xx2 + 3 = 23 `