Enumerating sums of integers So I found a question which seems to be really easy to answer but after thinking a lot about it, well, I've have to say, I came up with nothing.
So here's the question:

Assume the sum of $a+b+c+d+e+f$ that each of the numbers can be $0,3,4,5$, how many different summations can exist?

The only point I came up with is that the integer $0$ doesn't change the summation therefore different permutations of a $0$ count as one as I said the reason above.
And sorry if the Tags are irrelevant and my Math language is crappy I'm a newbie!
Please feel free to edit my tags I really do not know which one to add.
Any help would be highly appreciated in advance.
 A: Denote the set of all possible sums by $S$. Clearly $S\subset\{0,1,2,3,\cdots,30\}$.
Then take care of each element in $\{0,1,\cdots,30\}$.
$30=5+5+5+5+5+5$
$29=5+5+5+5+5+4$
$28=5+5+5+5+5+3$
$27=5+5+5+5+4+3$
$26=5+5+5+5+3+3$
Replacing one 5 by 0, we can obtain $25,24,23,22,21$. Repeat this process for 4 times until we get
$10=0+0+0+0+5+5$
$9=0+0+0+0+5+4$
$8=0+0+0+0+5+3$
$7=0+0+0+0+4+3$
$6=0+0+0+0+3+3$.
Then we see $S=\{0,3,4,5\}\cup\{6,7,\cdots,30\}$.
A: The smallest possible sum is $0+0+0+0+0+0=0$.  The largest is $5+5+5+5+5+5=30$.  It is also obvious that you can't have a sum of $1$ or $2$.  So, we have an upper bound of $29$ possible sums.  
Can you find a quick proof that all the numbers from $3$ to $29$ are possible sums?
A: The number of solutions is $29$.
The smallest sum that you can get is then $0+0+0+0+0+0=0$ and the highest is $5+\dots+5=30$. You can of course not get $1$ or $2$, so you have at most $29$ sums. It remains to find that you get exactly $29$ sums.
By adding only fives you can get $$\{5n\mid n=0,\dots 6\}.$$
By adding one $3$ or one $4$ and all others fives you can get $$\{5n+3\mid n=0,\dots 5\},$$ $$\{5n+4\mid n=0,\dots 5\}.$$
By choosing $3+3$ or $3+4$ you get
$$\{5n+6\mid n=0,\dots 4\},$$ 
$$\{5n+7\mid n=0,\dots 4\}.$$
This shows that you get $29$ distinct numbers.
