A two digit number is 4 times the sum of its digits

Let the require digit be $=(10 x+y)$

Then according to question

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

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

$10 x+y-4 x-4 y=0$

$6 x-3 y=0$

$2 x-y=0$

$2 x=y \ldots \ldots(1)$

And, $(10 x+y)=2 x y$......(2)

Now putting the value of y in equation (2) from (1)

$(10 x+2 x)=2 x \times 2 x$

$4 x^{2}-12 x=0$

$4 x(x-3)=0$

$x(x-3)=0$

So, either

$x=0$

Or

$(x-3)=0$

$x=3$

So, the digit can never be negative.

When $x=3$ then

$y=2 x$

$=2 \times 3$

$=6$

Therefore, number

$=10 x+y$

$=10 \times 3+6$

$=36$

Thus, the required number be 36