A Geometric Puzzle

With the numbers going from 1 to 30, there will be 30 distinct numbers, equally spaced around the circumference of the circle. In order for a number to be diametrically opposite to 7, it must be in-line with the diameter of the circle with one end on 7; this occurs when that number is half-way around the circle relative to 7. Since there are 30 spaces, half-way would be 15 spaces, and so we add 15 to 7:

7 + 15 = 22

which tells us that 22 is the number. I also decided to prove this geometrically/computationally. Using the tikz package in LaTeX, I was able to construct this problem with a grid line to further emphasize the opposite number to 7, which is of course 22.

Code:
    \begin{tikzpicture}
        \def \n {30}
        \def \radius {5}
        \draw circle(\radius)
              foreach \s in {1,...,\n}{
                  (-360/\n*\s+84:-\radius)
                  circle(.4pt) circle(.8pt) circle(1.2pt)
                  node[anchor=-360/\n*\s+84)]{$\s$}
              };
        \draw[help lines] (-5,-5) grid (5,5);
    \end{tikzpicture}

You could extend this puzzle for sure, either possible or impossible. To make it impossible, we can make the numbers run from 1 to an odd number, in which there are no opposites. I think there are some scenarios in which impossible puzzles may be appropriate; the act of finding out a puzzle is impossible could be viewed as solving it in a way, as it shows that the student understands the problem well enough to realize there is no solution.

A geometric puzzle can usually be solved using diagrams, whereas logical puzzles require more number theory.