Question

A two-digit number is such that the product of the digits is 12. when 9 is subtracted from the number, the digits are reversed

Answer 1

Let the digits be x and y.
No:10x+y
X*y=12
10x+y-9=10y+x
No:43

Answer 2

Given :

Product of the digits is 12.

When 9 is subtracted from the number, the digits are reversed.

To Find :

The number

Solution :

Let the unit digit in the number be x,

As the product of digits is 12, the ten's digit in the number is $\frac{12}{x}$

∴ The value of the number = 10 × $\frac{12}{x}$ + x

                                             = $\frac{120}{x}$ + x

When the digits are reversed, unit's digit becomes ten's digit and ten's digit becomes unit's digit and its value will become 10x + $\frac{12}{x}$

As per question,

  $\frac{120}{x}$ + x - 9             =  10x + $\frac{12}{x}$

⇒         9x               =   $\frac{120}{x} - \frac{12}{x} - 9$

⇒         9x               =  $\frac{120 - 12 - 9x}{x}$

⇒         9x²              =  108 - 9x

⇒    9x² + 9x - 108  =  0

⇒   9x² + (36 - 27)x - 108   = 0

⇒   9x² + 36x - 27x - 108   = 0

⇒ 9x(x + 4) - 27(x+4)          = 0

∴   (x + 4)(9x - 27)               = 0

Either x = -4 or x = 3

We can not have negative number in unit place

∴ 3 is in unit's place and $\frac{12}{3}$ i.e., 4 in the ten's place.

Therefore, the number is 43.