Number of different possible game scores on a quiz game with n questions On a quiz game, players get 5 points for a correct answer, 2 points for not answering, and 0 points for a wrong answer. If there are n questions, find a way to determine how many possible scores there are.

I'm thinking this isn't as hard as I think it is. All even numbers are a multiple of 2, and odd numbers ≥5 are given by 5 + multiples of 2. Would it be all integers between the maximum possible score (5n) and 0, discard scores 1 and 3? 
Even in this way, though, don't know how the 0 points for a wrong answer would be incorporated.

Edit: 1 and 3 don't work. I think 5n -1 , 5n - 2, 5n - 4 and 5n-7 don't work for a total of 6 solutions that don't work? So 5n - 6
On a quiz with $n$ questions, the score will be between 0 and $5n$ so there are at most $5n+1$ possibilities.

As observed above, scores of 1 and 3 are not possible at the ''bad'' end of the score range. If we instead consider the ''good'' end of possible high scores, we can observe the following. For any question not answered we lose 3 points, and for every incorrect answer we lose 5 points. Thus the scores
$$5n - 1,\, 5n-2,\, 5n-4,  \quad\text{and} \quad5n-7$$
are not possible, since the numbers $1, 2, 4,$ and $7$ cannot be written as a sum of $3$'s and $5$'s.

In all we have found 6 scores that are not possible: 2 in the ''bad'' range, and 4 in the ''good'' range. If we additionally assume that $5n - 7 &gt; 3$, i.e. $n &gt; 2$, then these are in fact 6 distinct scores that are not possible so there are at most $5n-5$ scores which are possible.

I'll make the claim that all remaining $5n-5$ scores are possible, and leave this proof as an exercise. 

Thus the answer is $5n-5$ when $n\geq 3$, and  the remaining cases can be computed separately: 6 possibilities with 2 questions and 3 possibilities with 1 question.
If you are looking for the number of possible scores as stated in your first paragraph rather than all the literal point values you could use combinations:
for the first 4 questions this formula translates simply to the combinations formula without repetition: 
$$
\frac{(r+n-1)!}{r!(n-1)!)}
$$
This formula is chosen rather than a permutation formula because the order in which you get different scoring possibilities does not matter, and you are able to get repetitions of the same score. note: that n in this example is not the same as "n" questions, I'll use "r" for questions.

To test if it worked later and to make it easier- turn the scoring n possibilities into variables:

5 points=A 
2 ponts = B 
0 points = C 
n=3 possibilities of scores, or n=3
there are 3 possibilities for every r questions. lets use two examples

EXAMPLE 1: 2 questions 
n=3
r=2
$$
\frac{(2+3-1)!}{2!(3-1)!)}=6
$$
you can test this by checking the combinations of points by using the assigned variables from above:
1: AA =10 points
2: AB, BA= 7 points
3: AC, CA =5 points
4: BB =4 points
5: BC, CB =2 points
6: CC =0 points
Note: even if you swap around these variables you still get the same points.. AB=BA=7 points

EXAMPLE 2: 4 questions 
n=3
r=4
$$
\frac{(4+3-1)!}{4!(3-1)!)}=15
$$

There are special cases:

for 5, 6, and 7 questions the possible scores are:
5 questions =20
6 questions = 25
7 questions = 30

because after 4 questions, the different combinations can end up resulting in the same number of points.

for every n>=4, the formula is:
5(n-1)