Graph theory restatement: Find the smallest $n$ such that a $K_{11}$ can be covered by some $K_n$s such that each vertex is included in at most $5$ of our $K_n$s. First lets try the case $n=3$, where we cover the $K_{11}$ with triangles. Since every vertex needs to be adjacent to $10$ other vertices and each triangle its in can satisfy $2$ of these adjacencies, every vertex needs to be in exactly $5$ triangles. This is, however, impossible, as $3\nmid 5\cdot 11$. The case $\boxed{n=4}$ is shown below and is quite weak. I've demonstrated that a $K_{13}$ can be covered by $K_4$s such that each vertex is included in at most $4$ of our $K_n$s.

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