5.85 It is known that a safe is managed by an 11-member committee, and several locks are added to the safe. The keys to these locks are distributed among the committee members. To ensure that any 6 members present can open the safe, but any 5 cannot, what is the minimum number of locks that should be added to the safe?

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[Solution] Let the minimum number of locks that meet the requirements be $n$. Denote this set of $n$ locks as $A$. Let $A_{i}$ be the set of locks that the $i$-th committee member can open. According to the given conditions, for any 5-element subset $\left\{i_{1}, i_{2}, \cdots, i_{5}\right\}$ of $\{1,2, \cdots, 11\}$, we have
$$
A_{i_{1}} \cup A_{i_{2}} \cup A_{i_{3}} \cup A_{i_{4}} \cup A_{i_{5}} \neq A ;
$$

For any 6-element subset $\left\{j_{1}, j_{2}, \cdots, j_{6}\right\}$ of $\{1,2, \cdots, 11\}$, we have
$$
A_{j_{1}} \cup A_{j_{2}} \cup \cdots \cup A_{j_{6}}=A .
$$

From (1), we know that the set $A-\bigcup_{k=1}^{5} A_{i_{k}}$ is non-empty. Let $x_{i_{1} \cdots i_{5}}$ be an element of this set. This indicates that $x_{i_{1} \cdots i_{5}}$ represents a lock that the 5 committee members with numbers $i_{1}, \cdots, i_{5}$ cannot open. Equation (2) implies that for any $j \notin\left\{i_{1}, i_{2}, \cdots, i_{5}\right\}$, we must have $x_{i_{1}, \cdots i_{5}} \in A_{j}$. In this way, we establish a correspondence between the 5-element subsets of $\{1,2, \cdots, 11\}$ and the locks.

We point out that this correspondence is injective. If not, suppose that for two different 5-element subsets $\left\{i_{1}, \cdots, i_{5}\right\}$ and $\left\{k_{1}, \cdots, k_{5}\right\}$, we have $x_{i_{1} \cdots i_{5}}=x_{k_{1} \cdots k_{5}}$. Since the two 5-element subsets are different, their union must have at least 6 elements, which would mean that 6 people cannot open the same lock, leading to a contradiction.

Since the number of different 5-element subsets of $\{1,2, \cdots, 11\}$ is $C_{11}^{5}=462$, the number of locks $n \geqslant 462$.

On the other hand, if we add 462 locks to the safe and correspond these locks one-to-one with the 462 5-element subsets of $\{1,2, \cdots, 11\}$. Distribute the 6 keys for each lock to the 6 committee members who are not in the 5-person group corresponding to that lock. Then, any 5 committee members will have one lock they cannot open, while any 6 committee members can open all the locks.
In summary, the minimum number of locks that should be added to the safe is 462.