How many possible combinations are there of these 4 numbers to add to multiples of 4? If I have the numbers 0 1 2 and 3, how many combinations of any size but maximum of 4 add to a multiple of 4?
EG:
0000 = 0 ( so 0*4), 


*

*1111 = 4,

*0000 = 0 ( so 0*4), 

*0112


How many combinations are there?
 A: As this is tagged homework, I will derive the solution for length $N=4$, which is the hardest part. But the framework is there to apply basically the same argument to any other length.
Treat the elements $\{0,1,2,3\}$ as elements to the group $G=(\mathbb{Z}_4,+)$.
The question is then "how many way can we pick exactly four elements $a,b,c,d$ from $G$, including the identity, such that $abcd=e$," with $e$ representing the identity.
If $abcd = e$, then $ab = (cd)^{-1}$. Consider $\alpha = ab $ and $\beta = (cd)^{-1}$. This induces the unique choices: $(\alpha,\beta) \in \{(0,0),(1,3),(2,2),(3,1)\}$. We see that the range of $\alpha$ is $G$. Thus, we can apply the argument again to $\alpha = ab$. Hence, for any choice of $\alpha$, there are exactly 4 ways to choose $a,b$ to generate $\alpha$.
Then, for any choice of $\alpha$, the value of $\beta$ is chosen for us. But again, we can choose $c,d$ in 4 different ways to obtain $\beta$.
Hence, there are $4\times 4\times 4 = 4^3 = 64$ ways to add $0,1,2,3$ to get to a multiple of $4$.

Some Python code to verify:
>>>a=0
>>>for i in range(0,4):
       for j in range(0,4):
           for k in range(0,4):
               for l in range(0,4):
                   s = i+j+k+l
                   if s % 4 == 0:
                       a = a+1
>>> a
64

A: For length $4$. Suppose you already picked the first three. Then there is exactly one number that you can put in the end to make it a multiple of 4. Therefore there are $4^3=64$ such lists.
Using the same method we see the total is $4^0+4^1+4^2+4^3$=$\frac{4^4-1}{4-1}=\frac{255}{3}=85$
