Function continuous at odd numbers from $0$ to $99$, but discontinuous everywhere else. I have an idea about this but am not entirely sure if it's right. 
If I have a function, say $g(x) = (x-1)(x-3)(x-5)...(x-99)$
The roots of g(x) are only odd numbers from 0 to 99. 
$$
f(x) = \begin{cases} 
      g(x) & x \in\mathbb{Q} \\
      0 & x \notin\mathbb{Q}
   \end{cases}
$$
Is this correct?

Your approach is correct. For any odd number $x$ between $0$ and $99$,
$$ \lim_{y \to x} |f(y)| \leq \lim_{y \to x} |g(y)| = 0 = f(x)
$$
Otherwise $f$ locally behaves like the indicator function of $\mathbb{Q}$, which is discontinuous.
