6.1. Solve the equation $\frac{n!}{2}=k!+l!$ in natural numbers, where $n!=1 \cdot 2 \ldots n$. If there are no solutions, indicate 0; if there is one solution, indicate $n$; if there are multiple solutions, indicate the sum of the values of $n$ for all solutions. Recall that a solution is a triplet ( $n, k, l$); if solutions differ in at least one component, they are considered different.
Answer: 10 (all triples of solutions $(n, k, l):(3,1,2),(3,2,1),(4,3,3)$).

Solution. Note that if $k4$, then $n!>4 \cdot(n-1)!\geqslant 2 \cdot(k!+l!)$, so there are no such solutions. If $n=2$, we get $1=k!+l!$ - no solutions; if $n=3$, the equation $3=k!+l!$ has two solutions: $k=1, l=2$ and $k=2, l=1$; if $n=4$, the equation $12=k!+l!$ gives one more solution $k=l=3$.