Which two digit number when you find the product of the digits yields a number that is half the original? Which two digit number when you find the product of the digits yields a number that is half the original?
Let x=$ab$ be the $2$-digit number. So $x=10a+b$.
Then $ab=\frac{x}{2} \implies ab=\frac{10a+b}{2} \implies 2ab=10a+b \implies b=\frac{10a}{2a-1}$. I guess $a=3$ and get $b=6$. So the answer is $36$.
But how can this be done without guessing.

$2ab-10a-b=0$ is a simple quadratic Diophantine equation. The canonical approach in this case is the following.
You add $5$ to both sides and you obtain
$$
2ab-10a-b+5=5,\ \ \Rightarrow (2a-1)(b-5)=5
$$
so $b-5$ or $2a-1$ must be equal to $5$ (or $-5$, but it is not possible) and the other must be equal to $1$.
But $0<b<10$ so $b-5$ can't be $5$, it must be $1$, hence $b=6$ and thus $2a-1=5$, from which $a=3$.
