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

Let tens digit be x and units digit be y.

Given,

$xy = 18 \\[1em] \Rightarrow y = \dfrac{18}{x}$

The number = 10x + y = $10 \times x + \dfrac{18}{x} = \dfrac{10x^2 + 18}{x}$

Reversed number = 10y + x = $10 \times \dfrac{18}{x} + x = \dfrac{180 + x^2}{x}$

Given,
if 63 is subtracted from the number, the digits are reserved.

$\Rightarrow \dfrac{10x^2 + 18}{x} - 63 = \dfrac{180 + x^2}{x}\\[1em] \Rightarrow \dfrac{10x^2 + 18 - 63x}{x} = \dfrac{180 + x^2}{x}\\[1em] \Rightarrow 10x^2 + 18 - 63x = 180 + x^2\\[1em] \Rightarrow 10x^2 + 18 - 63x - 180 - x^2 = 0\\[1em] \Rightarrow 9x^2 - 63x - 162 = 0\\[1em] \Rightarrow x^2 - 7x - 18 = 0\\[1em] \Rightarrow x^2 - 9x + 2x - 18 = 0\\[1em] \Rightarrow x(x - 9) + 2(x - 9) = 0\\[1em] \Rightarrow (x - 9)(x + 2) = 0\\[1em] \Rightarrow (x - 9) = 0 \text{ or } (x + 2) = 0\\[1em] \Rightarrow x = 9 \text{ or } x = - 2$

As the digit of a number cannot be negative. So x = 9.

y = $\dfrac{18}{x} = \dfrac{18}{9}$ = 2

The number = 10x + y = 10 $\times$ 9 + 2 = 90 + 2 = 92

Hence, the number = 92.