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  ` 10 y + x`.

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

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

` ⇒ 10 y + x = 4x + 4 y `

`⇒ 4 x + 4 y - 10 y - x = 0 `

`⇒ 3 x - 6 y = 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

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

`⇒ 10 x + 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 systems of equations for xand 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 xx2=0`

` ⇒ x - 4 =0`

` ⇒ x = 4`

Hence, the number is ` 10 xx2 + 4 = 24`.