13. In a store, there are large, medium, and small boxes of marbles, containing 13, 11, and 7 marbles respectively. If someone wants to buy 20 marbles, they don't have to break open the boxes (1 large box plus 1 small box). If someone wants to buy 23 marbles, they must break open the boxes. Please find the smallest number such that anyone buying more than this number of marbles definitely doesn't have to break open the boxes. This smallest number is ( ).
【Analysis】Extremum and reasoning problem. Since 23 is an odd number, an even number of boxes is required. 2 boxes are clearly insufficient, 6 boxes will definitely exceed, and with 4 boxes, $28<30$, so 30 cannot be achieved.
$$
\begin{array}{l}
31=7+11+13 ; \quad 32=7+7+7+11 ; \\
33=7+13+13 ; \quad 34=7+7+7+13 ; \\
35=7+7+7+7+7 ; \quad 36=7+7+11+11 ;
\end{array}
$$
$37=11+13+13$, these 7 consecutive natural numbers can all be achieved,
and the small box is 7, so as long as some more small boxes are added, all numbers above 31 can be achieved, so the required number is 30.