Question

A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Answer 1

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

The number is 4 times the sum of the two digits. Thus, we have

`10y + x = 4(x + y)`

`⇒ 10y + x = 4x + 4y`

`⇒ 4x + 4y - 10y - x = 0`

`⇒ 3x - 6y = 0`

`⇒ 3(x - 2y) = 0`

`⇒ x - 2y = 0`

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

If 18 is added to the number, the digits are reversed. Thus, we have

`(10y + x) + 18 = 10x + y`

`⇒ 10x + y - 10y - x = 18`

`⇒ 9x - 9y = 18`

`⇒ 9(x - y) = 18`

`⇒ x - y = 18/9`

`⇒ x - y = 2`

So, we have the systems of equations

`x - 2y = 0`

`x - y = 2`

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

Subtracting the first equation from the second, we have

`(x - y) - (x - 2y) = 2 - 0`

`⇒ x - y - x + 2y = 2`

`⇒ y = 2`

Substituting the value of y in the first equation, we have

`x - 2 xx 2 = 0`

`⇒ x - 4 = 0`

`⇒ x = 4`

Hence, the number is `10 xx 2 + 4 = 24`.