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.

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

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.