Question

A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Answer 1

Let the tens and the units digits of the required number be x and y, respectively. 

Then, we have: 

xy = 18 …….(i) 

Required number = (10x + y) 

Number obtained on reversing its digits = (10y + x) 

∴(10x + y) - 63 = 10y + x 

⇒9x – 9y = 63 

⇒ 9(x – y) = 63 

⇒ x – y = 7 ……..(ii) 

We know: 

(x + y)² – (x – y)² = 4xy

⇒ (x + y) = ± √(x − y)² + 4xy 

⇒ (x + y) = ± √49 + 4 ×18 

 = ± √49 + 72 

 = ± √121 = ±11 

⇒ x + y = 11 ……..(iii) (∵ x and y cannot be negative) 

On adding (ii) and (iii), we get:

2x = 7 +11 = 18 

⇒x = 9 

On substituting x = 9 in (ii) we get 

9 – y = 7 

⇒ y = (9 – 7) = 2 

∴ Number = (10x + y) = 10 × 9 + 2 = 90 + 2 = 92 

Hence, the required number is 92.