A two-digit number contains the smaller of the two digits in the unit place. The product of the digits is 40 and the difference between the digits is 3. 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 = 40

⇒ xy = 40     ………(1)

Given,

Difference between digits = 3

⇒ x - y = 3

⇒ x = y + 3     ………(2)

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

⇒ (y + 3)y = 40

⇒ y² + 3y - 40 = 0

⇒ y² + 8y - 5y - 40 = 0

⇒ y(y + 8) - 5(y + 8) = 0

⇒ (y - 5)(y + 8) = 0

⇒ (y - 5) = 0 or (y + 8) = 0     [Using zero-product rule]

⇒ y = 5 or y = -8

Since, digits should be positive, thus y ≠ -8.

⇒ x = y + 3 = 5 + 3 = 8.

The number = 10x + y = 10(8) + 5 = 85.

Hence, the required number is 85.