Hey Alan, thanks for your feedback! If you need to ask a question about anything that goes on on the ZIML site (Daily Magic Spells, Monthly Comptetition, etc.) feel free to start a new topic for discussion on the forum :)

Note on Dec 10 the problem is asking for the probability that the number will "not increase" the median, so it is ok if it stays the same or if it decreases, hence the $6$ possible values.

On the problem on Dec 8, since we know $x$ and $y$ satisfy a linear equation, we can write $x^2-y^2$ in terms of only $x$, which yields the quadratic function $-0.096x^2 + 31.92x -129.96$. This is a parabola that opens downwards, and thus has a global maximum. That is, the sum cannot grow indefinitely. You are however, correct, that the problem should ask for the values of $x$ and $y$ to be both positive, otherwise the presented solution is incorrect. Note the parabola $-0.096x^2 + 31.92x -129.96$ has a global maximum when $x = 16.625$, so the maximum value of $x^2 - y^2$ will occur when $x$ is as close as possible to $16.625$. All $x\equiv 1 \pmod{5}$ yield an integer solution to the equation, so $x = 16$ is the closest one. When $x = 16$, $y = -11$, so $16^2 - (-11)^2 = 135$ is the maximum possible value for $x$ and $y$ integers, and $6^2 - 3^2 = 27$ is the maximum value for $x$ and $y$ positive integers.