A two-digit number is 4 times the sum of its digits and twice the product of its digit

Solution:

Let the digits at units and tens places be $x$ and $y$, respectively.

$\therefore$ Original number $=10 y+x$

According to the question:

$10 y+x=4(x+y)$

$\Rightarrow 10 y+x=4 x+4 y$

$\Rightarrow 3 x-6 y=0$

$\Rightarrow 3 x=6 y$

$\Rightarrow x=2 y \ldots .(\mathrm{i})$

Also,

$10 y+x=2 x y$

$\Rightarrow 10 y+2 y=2.2 y . y \quad[$ From (i) $]$

$\Rightarrow 12 y=4 y^{2}$

$\Rightarrow y=3$

From (i), we get:

$x=2 \times 3=6$

$\therefore$ Original number $=10 \times 3+6=36$