A number consists of two digits whose product is 18. If 27 is added to the number, the digits interchange their places. Find the number.

Let the ten's and unit's digits of required number be x and y respectively.

Number = 10x + y

Given,

Product of digits = 18.

⇒ xy = 18

⇒ x = $\dfrac{18}{y}$     ………(1)

Given,

If 27 is added to the number, the digits interchange the place.

⇒ 27 + 10x + y = 10y + x

⇒ 27 + 10x + y - 10y - x = 0

⇒ 9x - 9y + 27 = 0

⇒ 9(x - y + 3) = 0

⇒ x - y + 3 = 0

⇒ x - y = -3     ………(2)

Substituting the value of x from equation (1) in equation (2),

⇒ $\dfrac{18}{y}$ - y = -3

⇒ $\dfrac{18 - y^2}{y}$ = -3

⇒ 18 - y² = -3y

⇒ y² - 3y - 18 = 0

⇒ y² - 6y + 3y - 18 = 0

⇒ y(y - 6) + 3(y - 6) = 0

⇒ (y + 3)(y - 6) = 0

⇒ (y + 3) = 0 or (y - 6) = 0     [Using zero-product rule]

⇒ y = -3 or y = 6.

Since, digits must be positive y ≠-3.

Substituting value of y = 6 in equation (1), we get :

x = $\dfrac{18}{6}$ = 3.

The number is : 10 × 3 + 6 = 36.

Hence, the required number is 36.