Question

A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from 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 more than 6 times the sum of the two digits. Thus, we have

` 10 y + x = 6 (x+y)+4`

` ⇒ 10y +x =6x + 6y + 4`

`⇒ 6x + 6y -10y -x=-4 `

` ⇒ 5x -5y =-4`

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

If 18 is subtracted from the number, the digits are reversed. Thus, we have

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

`⇒ 10x + y -10y -x = -18 `

` ⇒ 9x -9y =-18`

` ⇒ x -y =-18/9`

` ⇒ x - y = -2`

So, we have the systems of equations

` 5x - 4y = -4 `

` x - y =-2`

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

Multiplying the second equation by 5 and then subtracting from the first, we have

`(5x-4y)-(5x-5y)=-4-(-2xx5)`

` ⇒ 5 x -4y -5x +5y =-4+10`

` ⇒ y = 6`

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

` x - 6=-2`

`⇒ x = 6-2 `

` ⇒ x =4`

Hence, the number is `10 xx6+4=64.`