Problem 193

Let Mary's score, number correct, and number wrong be $s,c,w$ respectively. Then
\begin{align*} s&=30+4c-w \\ &=30+4(c-1)-(w-4) \\ &=30+4(c+1)-(w+4). \end{align*}
Therefore, Mary could not have left at least five blank; otherwise, one more correct and four more wrong would produce the same score. Similarly, Mary could not have answered at least four wrong (clearly Mary answered at least one right to have a score above $80$, or even $30$.)
It follows that $c+w\geq 26$ and $w\leq 3$, so $c\geq 23$ and $s=30+4c-w\geq 30+4(23)-3=119$. So Mary scored at least $119$. To see that no result other than $23$ right/$3$ wrong produces $119$, note that $s=119\Rightarrow 4c-w=89$ so $w\equiv 3\pmod{4}$. But if $w=3$, then $c=23$, which was the result given; otherwise $w\geq 7$ and $c\geq 24$, but this implies at least $31$ questions, a contradiction. This makes the minimum score $\boxed{119}$.