Finding integer solutions out of a a | b Determine all positive integer values of (n) such that
$$ { n \choose 0 } + { n \choose 1 } + { n \choose 2 } + { n \choose 3 } \ \bigg| \ 2 ^ { 2008 } $$
What is the sum of all these values? 
CURRENT PROGRESS:
I was able to find out that this is equivalent to $(n+1)(n^2 - n + 6) \,| \,3\times (2)^{2009}$  and opened in 2 cases, $n+1 = 2^a$ and $n+1 = 3\times 2^a$, trying to solve like: $n = 2^a - 1$, then $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8$, doing the same to the 2nd case but couldn't find solutions. Something that should be useful is that $ n^2 - n + 6 = 2^{2a} - 3\times2^a + 8 | \space\space 3\times2^{2019-a}$, it has a factor 3 in it.

So $(n+1)(n^2 -n + 6) = 3\cdot 2^{2009}$ so 
So $n+1 = 3^t2^s; n^2 -n+6 = 3^r2^w; t+r \le 1; s + w \le 2009$.
Case 1:$t= 0$.
$n = 2^s -1$ and $n^2 - n + 6 = 2^{2s} - 2^{s+1} -2^s + 6$.
If $s = 0$ we have $n=0$ and that's a solution (if we assume ${0\choose k}=0$ for $k \ne 0$). If $s \ge 1$ then
$n^2 -n +6 = 2(2^{2s-1} -2^s - 2^{s-1} + 3)$. 
If $s = 1$ we have $n=1$ and that's a solution (if we assume ${n\choose k} = 0$ if $n > k$).  If $s \ge 1$ then
$2^{2s-1} - 2^s - 2^{s-1} + 3$ is odd so either is either equal to $3$ or $1$.
So either $2^{2s-1}=  2^s + 2^{s-1}$
$2^{s} = 2 + 1$ which is impossible, or
$2^{2s-1}+2 = 2^s + 2^{s-1}$ so
$2^{2s-2} + 1 = 2^s + 2^{s-2}$ so $s-2 =0$ and $s = 2s -2$ or $s=2$ so $n= 3$ and we have $n+1 = 4$ and $n^2-n+6 = 9$.  That's a solution.
So far: $n = 0,n=1, n=3$ are solutions
Case 2: $t = 1$
$n = 3*2^s - 1$ and $n^2 - n + 6 = 9*2^{2s} - 3*2^{s+1} + 1 - 3*2^{s} - 1 + 6 = 3(3*2^{2s} - 2^{s+1} + 2^s + 2) = 2^w$.
That's impossible.
So unless I made a mistake the only solutions are $0, 1,3$ and if we don't consider ${n< k \choose k}$ a legitimate value then the only solution is $n=3$.
