13. Determine the maximum positive integer $k$ such that $k^{2}$ divides $\frac{n !}{(n-6) !}$ for every $n>6$.


13. The answer is 12. Let $N=\frac{n !}{(n-6) !}=(n-5)(n-4)(n-3)(n-2)(n-1) n$.

First, let's find the maximum $q$ such that $q$ divides $N, \forall n$. Among six consecutive integers, three are divisible by 2, and of these three, one is certainly divisible by 4. Therefore, $2^{4}$ divides $N$. Among six consecutive integers, there are two integers divisible by 3. Therefore, $3^{2}$ divides $N$. Among six consecutive integers, only one is divisible by 5. Therefore, 5 divides $N$. Thus,

$$
q \geq 2^{4} \cdot 3^{2} \cdot 5
$$

Since we are looking for the maximum $q$ such that $q$ divides $N, \forall n$, consider $n_{1}=7, n_{2}=13$. Certainly, $q$ divides both $N_{1}$ and $N_{2}$, so we have that

$$
q \text { divides } \operatorname{GCD}\left(N_{1}, N_{2}\right)
$$

We find that $N_{1}=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7, N_{2}=8 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13$. Therefore, $\operatorname{GCD}\left(N_{1}, N_{2}\right)=2^{4} \cdot 3^{2} \cdot 5$. Now, from 1 and 2, it follows that $q=2^{4} \cdot 3^{2} \cdot 5$. To extract the maximum square from $q$, we find that $k^{2}=2^{4} \cdot 3^{2}$, which means $k=12$.