Simple but deadly

As Dylon points out, we can use the fact that in a finite group the order of any element divides the order of the group, a corollary to Lagrange's Theorem.

For a fixed positive integer $n$ , let $G$ be the multiplicative group modulo $2^n-1$ , with $ord(G)=\phi(2^n-1)$ . Now $ord(2)=n$ , so that $n|\phi(2^n-1)$ . Letting $n=2015$ , we see that the answer is $\boxed{0.}$

Consider $U_{2^{2015}-1}$ , the set of all residues in modulo $2^{2015}-1$ relatively prime to $2^{2015}-1$ . Being in this set implies that an element is a 'unit', or has a multiplicative inverse.

Note that $2 \in U_{2^{2015}-1}$ , and the order of two is $2015$ . I'll sketch out a quick proof that $a \in U_k \Rightarrow ord(a)|\varphi(k)$ :

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Observe that $a^{\varphi(k)}\equiv1\pmod{k}$ , which is just Euler's totient theorem. Now, consider the following set of powers of $a$ :

$\{a^1,a^2,a^3 ... a^{ord(a)} \}$

Since $a^{ord(a)}\equiv1,$ we have $a^k\equiv a^{k \pmod{ord(a)}}$ , right? So since by definition for $k<ord(a)$ we have $a^k\neq1$ we know $a^k\equiv1$ iff $k$ is $0 \pmod{ord(a)}$ or $ord(a)|k$ . Hence, $ord(a)|\varphi(k)$ .

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Now we done, since $ord(2)|\varphi(2^{2015}-1)$ or $2015|\varphi(2^{2015}-1)$ .