The answer is $ d(2008) = 8$. Let $ a \in X$ be an element of $ X$. Then using the definition of $ X$ it is not hard to prove inductively that $ ka \in X$ iff $ k$ is not divisible by $ 3$. (First prove that $ 2a \in X$, then that $ 3a \notin X$, then that $ 4a \in X$, and so on.) Then letting $ x\in X$ be the minimal element of $ X$ suppose there is some $ l\in X$ which is not divisible by $ x$. Applying our previous considerations to the multiples of $ x$ and $ l$ we see $ x$ and $ l$ must be divisible by the same power of $ 3$ (otherwise some multiple of $ 3x$, say, would be in $ X$). Choose minimal $ l\in X$ not divisible by $ x$ (assuming such exists). Then both $ |l - x|, |l - 2x| < l$ are not divisible by $ x$ and can't be in $ X$ since they are $ < l$. Hence $ l + x, l + 2x \in X$. But one of these numbers then is divisible by a higher power of $ 3$ than $ x$ and we get a contradiction. Thus $ X$ consists only of multiples of $ kx$ for which $ k$ is not divisible by $ 3$. Since $ 2008$ is not divisible by $ 3$ the answer follows.

Xixas wrote: First prove that $ 2a \in X$ How? Sorry if I appear stupid

nuclear_alchemist wrote: Xixas wrote: First prove that $ 2a \in X$ How? Sorry if I appear stupid Take $ a = b \in X$, as $ |a - b| = 0$ is not a positive integer and thus not in $ X$, we must have $ a + a = 2a \in X$.

for x=y=a one of a+a and |a-a| should be there in the set but it is a set of +ve number so |a-a| cannot.

Thank you!