On the plane, $n$ lines are drawn such that every two intersect, but no four pass through the same point. There are a total of 16 intersection points, and through 6 of them, three lines pass. Find $n$.

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"Move" the given construction in such a way that any two lines still intersect, but no three lines pass through the same point. Then, if some three lines intersected at a certain point $O$, now instead of one point $O$, there will be three points of pairwise intersection of these lines. Therefore, as a result of "moving" the initial number of lines will not change, but the number of intersection points will increase by $2 \cdot 6=$ 12. In the end, all lines will intersect pairwise, and the number of intersection points will become $16+12=28$. Thus, $1 / 2 n(n-1)=28$, from which $n=8$.

## Answer

$n=8$.