Looking at Digits

Some sneaky factorisation works here: we can rewrite the given condition $2AB=10A+B$

in the form $(2A-1)(B-5)=5$

From this it's easy to see the only solution (in digits $A,B$ ) is $A=3$ , $B=6$ giving the answer $\boxed9$ .

$10A+B=2AB\implies B=$

$\dfrac{10A}{2A-1}=5+\dfrac{5}{2A-1}$ .

Since $B\leq 9, \dfrac{5}{2A-1}\leq 4\implies A\geq 2$ .

Since $B$ is an integer, $\dfrac{5}{2A-1}\geq 1\implies A\leq 3$ .

Hence $A$ can either be $2$ or be $3$ . It's easy to check that $B$ is an integer for $A=3$ only, and the value of $B$ is $6$ .

So $A+B=\boxed 9$ .

$\large \begin{aligned} AB = & \space 2 × A × B \\ 10x+y = & \space 2xy\end{aligned}$

If $2xy$ ends with the digit $y$ (as seen in $10x+y$ ), then $2x$ must either be $1$ or $6$ . But $2x$ can't be $1$ since it would mean that $x$ is non-integer. So $x$ must be $3$ .

$\large 30+y = 6y$

The only multiples of $6$ in the line of $30$ is $30$ and $36$ . If $30+y=30$ , that would mean $y$ is $0$ . So $30+y$ must be equal to $36$ . Meaning, $y$ must be $6$ .

$\large \begin{array}{ccc} A = 3, & & B = 6 \end{array}$

The values' sum is $3+6=\boxed{9}$ .