Question

A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digit interchange their places. Find the number. 

Answer 1

Let the digits at units and tens places be x and y, respectively. 

∴ xy = 14 

⇒ y = `14/x`   ...(1) 

According to the question: 

(10y + x) + 45 = 10x + y 

⇒ 9y – 9x = –45 

⇒ y – x = 5   ...(2) 

From (1) and (2), we get 

`14/x-x=-5` 

⇒ `(14-x^2)/x=-5` 

⇒ 14 – x² = –5x 

⇒ x² – 5x – 14 = 0 

⇒ x² – (7 – 2)x – 14 = 0 

⇒ x² – 7x + 2x – 14 = 0 

⇒ x(x – 7) + 2(x – 7) = 0 

⇒ (x – 7) (x + 2) = 0 

⇒ x – 7 = 0 or  x + 2 = 0 

⇒ x = 7 or x = –2 

⇒ x = 7   ...(∵ The digit cannot be negative) 

Putting x = 7 in equation (1), we get 

y = 2 

∴ Required number = 10 × 2 + 7

= 27 

Answer 2

Let the tens digit be x and ones digit be y.

Given, xy = 14   ...(i)

The number is 10x + y.

Given, that when 45 is added to the number the digits get interchanged red.

Hence, 10x + y + 45 = 10y + x

⇒ 9x – 9y + 45 = 0

⇒ x – y + 5 = 0   ...(ii)

From equations (i) and (ii), we get

`x - 14/x + 5 = 0`

⇒ x² – 14 + 5x = 0

⇒ x² + 5x – 14 = 0

⇒ x² + 7x – 2x – 14 = 0

⇒ x(x + 7) – 2(x + 7) = 0

⇒ (x + 7)(x – 2) = 0

⇒ x + 7 = 0 and x – 2 = 0

⇒ x = –7 and x = 2

Since, the digits cannot be negative, x = 2

Thus, `y = 14/x`

= `14/2`

= 7

Therefore, the number is (10x + y) = 27.