jmc

We can rewrite $A{A}_5$ and $B{B}_7$ to get $$\begin{array}{rl}5A+A& =7B+B\Rightarrow \\ 6A& =8B\Rightarrow \\ 3A& =4B.\end{array}$$We can see that the smallest possible values for $A$ and $B$ are $A=4$ and $B=3$. So the integer can be expressed as $44_5=33_7=\boxed{24_{10}}$.