A two-digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. 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 `10y + x`.

The product of the two digits of the number is 20. Thus, we have ` xy =20`

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

If 9 is added to the number, the digits interchange their places. Thus, we have

`(10y+x)+9=10x+y`

`⇒ 10y +x +9 =10x +y`

` ⇒ 1=x + y -10y -x =9`

` ⇒ 9x -9y =9`

`⇒ 9( x - y)=9`

` ⇒ x -y 9/9`

` ⇒ x - y =1`

So, we have the systems of equations

`xy=20`

`x-y=1`

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

Substituting `x =1+y` from the second equation to the first equation, we get

`(1+y )y=20`

`⇒ y + y^2 = 20`

`⇒ y^2 + y = 20 =0`

` ⇒ y^2 + 5y -4y -20 =0`

`⇒ y (y + 5)-4(4+5)=0`

`⇒ ( y + 5)(y - 4)=0`

`⇒ y = -5` Or `y = 4`

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

Hence, the number is `10 xx4 + 5= 45`.

Note that in the first pair of solution the values of *x* and *y* are both negative. But, the digits of the number can’t be negative. So, we must remove this pair.