A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let the ten's and unit digit be x and \frac{8}{x} respectively \\ Then, (10 x + \frac{8}{x}) + 18 = 10 \times \frac{8}{x} + x \\ \Rightarrow 10 x^{2} + 8 + 18 x = 80 + x^{2} \\ \Rightarrow 9 x^{2} + 18 x - 72 = 0 \\ \Rightarrow x^{2} + 2 x - 8 = 0 \\ \Rightarrow (x + 4) (x - 2) = 0 \\ \Rightarrow x = 2 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{ten's}}\,{\text{and}}\,{\text{unit}}\,{\text{digit}}\,{\text{be}}\,x\,{\text{and}}\,\frac{8}{x}\,{\text{respectively}} \cr & {\text{Then}},\left( {10x + \frac{8}{x}} \right) + 18 = 10 \times \frac{8}{x} + x \cr & \Rightarrow 10{x^2} + 8 + 18x = 80 + {x^2} \cr & \Rightarrow 9{x^2} + 18x - 72 = 0 \cr & \Rightarrow {x^2} + 2x - 8 = 0 \cr & \Rightarrow \left( {x + 4} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2 \cr}$