A two digit positive number is such that the product of its digit is 6. If 9 is added to the number , the digits interchange their place . Find the number.

Let the digit at unit's place be x .

Since, the product of digits is 6 , it's ten's digit = $\dfrac{6}{x}$

$\therefore \text{ Number } = 10 \times \dfrac{6}{x} + x \\[0.5em] = \dfrac{60}{x} + x \\[0.5em] = \dfrac{60 + x^2}{x} \\[0.5em] = \dfrac{x^2 + 60}{x} \\[0.5em]$

Given, if 9 is added to the number , the digits interchange their place

On interchanging the digits, number becomes = $10 \times x + \dfrac{6}{x}$

Acccording to given,

$\Rightarrow \dfrac{x^2 + 60}{x} + 9 = 10 \times x + \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2 + 9x + 60}{x} = 10x + \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2 + 9x + 60}{x} = \dfrac{10x^2 + 6}{x} \\[1em] \Rightarrow x^2 + 9x + 60 =10x^2 + 6 \\[1em] \Rightarrow 10x^2 - x^2 - 9x + 6 - 60 = 0 \\[1em] \Rightarrow 9x^2 - 9x - 54 = 0 \\[1em] \Rightarrow 9(x^2 - x - 6) = 0 \\[1em] \Rightarrow x^2 - 3x + 2x - 6 = 0 \\[1em] \Rightarrow x(x - 3) + 2(x - 3) = 0 \\[1em] \Rightarrow (x + 2)(x - 3) = 0 \\[1em] \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = 3$

Since the number is positive hence x ≠ -2.

If x = 3 ,

$\text{ Number } = \dfrac{x^2 + 60}{x} \\[1em] = \dfrac{3^2 + 60}{3} \\[1em] = \dfrac{69}{3} \\[1em] = 23$

Hence, the required number is 23.