Peculiar nature of the different marks obtainable in a test 
In an examination of $20$ questions, a student gets $-1,0,4$ marks for incorrect, unattempted, and correct answers respectively. Let $S$ be set of all the marks that a student can score. What is the number of distinct elements in $S$?

In the range of $101$ marks $-20,-19,\cdots80$, the marks $\{+79,+78,+77,+74,+73,+69\}$ are unobtainable, leaving us $95$ elements in $S$.
I find this rather strange. What is so special about this set of marks that makes them unobtainable? I couldn't find any pattern in them, but they are very close to each other, lying around $70$.
 A: From a perfect score of $80$, for every unmarked or wrong question, $4$ or $5$ points are lost.
$20$ correct $\to $ score $\in \{80\}$.
$19$ correct $\to $ score $\in \{75,76\}$
$18$ correct $\to $ score $\in \{70,71,72\}$
$17$ correct $\to $ score $\in \{65,66,67,68\}$
$16$ correct $\to $ score $\in\{60,61,62,63,64\}$
And from now on all lower numbers will be achieved.
A: These are the gaps between $\{+80\}$ (the score with $20$ questions right), $\{+76, +75\}$ (the possible scores with $19$ questions right), $\{+72, +71, +70\}$ (the possible scores with $18$ questions right), and $\{+68, +67, +66, +65\}$ (the possible scores with $17$ questions right).
When fewer than $17$ questions are correct, the incorrect/blank scores give enough flexibility that no gaps are left.
A: I think that following the layout of the possible scores in @paw88789's answer we see that what is "special" about these answers is not that they are close to 70 but that they are close to 80.
I think it's possibly helpful to view an equivalent problem. Where scores are given as 0,4 or 5. Here we would have the unobtainable scores as $$\{1,2,3,6,7,11\} \\ =\{80-79,80-78,80-77,80-74,80-73,80-69\}$$ To me, this is more intuitive as to why these are unobtainable. They cannot be reached with only 4s and 5s. While scores beyond 12 are always obtainable as linear combination of 4 and 5 (with positive coefficients).
A: We can recast the situation as being that there's a maximum score of $80$, and one can lose $5,4,$ or $0$ on each question (If you skip a question, you're giving up a possible $4$ points. If you answer a question incorrectly, you're not only losing one point, but also giving up an opportunity for $4$ points, so you're losing out on a total of $5$ points). Then your listed impossible marks correspond to losing $1,2,3,6,7,$ or $11$ points. This now looks like the Chicken Nugget problem. Using the $mn-m-n$ formula, we get $20-5-4=11$. If $m=5$, $n=4$, and $k$ is an impossible number, we can organize these impossible numbers as
$k<n: 1,2,3$
$m<k<2n: 6,7$
$2m<k<3n: 11$
If we try to continue this pattern, we get $3m<k<4n$, but there are no numbers between $3m$ and $4n$.
