13. A and B are two car retailers (hereinafter referred to as A and B) who ordered a batch of cars from a car manufacturer. Initially, the number of cars A ordered was 3 times the number of cars B ordered. Later, due to some reason, A transferred 6 cars from its order to B. When picking up the cars, the manufacturer provided 6 fewer cars than the total number ordered by A and B. In the end, the number of cars A purchased was twice the number of cars B purchased. Try to find: what is the maximum number of cars that A and B finally purchased in total? And what is the minimum number of cars they purchased in total?
Three, 13. Let the total number of cars finally purchased by A and B be $x$ units. In the 6 fewer cars supplied by the factory, A requested $y$ fewer cars $(0 \leqslant y \leqslant 6)$, and B requested $(6-y)$ fewer cars. Then we have
$$
\begin{array}{l}
\frac{3}{4}(x+6)-6-y \\
=2\left[\frac{1}{4}(x+6)+6-(6-y)\right] .
\end{array}
$$

After rearranging, we get $x=18+12 y$.
When $y=6$, $x$ is at most 90;
When $y=0$, $x$ is at least 18.
Therefore, the total number of cars purchased by A and B is at most 90 units and at least 18 units.