Locks and keys so any four of seven can open a safe

Seven employees share a safe fitted with several locks, all of which must be opened. Keys are distributed so that any $4$ employees together can open the safe, but no group of $3$ can. If $n$ is the minimum number of locks and $m$ is the number of keys each employee then holds, compute $10m+n$.

235. Every trio must be blocked by some lock none of them can open, and two different trios cannot share that lock (their union has at least $4$ people, who must open it). So $n=\binom73=35$, each lock's keys go to the $4$ people outside its trio, and by symmetry each person holds $35\cdot4/7=20$ keys. Thus $10\cdot20+35=235$.