<s>[INST] 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.
 [/INST] Set $n+1=m$
$n^2-n+6=(m-1)^2-(m-1)+6=m^2-3m+8$
$g=(m, m^2-3m+8)=(m,8)$
Check all the four possible cases
For example,
If $g=4$
$\dfrac{n+1}4\cdot\dfrac{ n^2-n+6}4$ will divide $3\cdot2^{2008-4}$
where the two factors are relatively prime
i.e. one of the factors must be odd $1$ or $3$
</s>