A lot of $1$ s

$x = \dfrac{10^{2015}-1}{9} = \underbrace{11111\ldots111}_{2015 \, \, \text{digits}}$

We know that the prime factorization of $11111 = y$ is $41 \times 271$ , which means that it has $(1+1)(1+1) = 4$ divisors or factors.

Since the number of digits of $x$ is a multiple of the number of digits of $y$ , we can write $x$ as $y+10^5y+10^{10}y+\ldots+10^{2010}y = y(1+10^5+10^{10}+\ldots+10^{2010})$ which shows us that $\boxed{x \, \, \text{is a composite number with more than} \, \, 4 \, \, \text{factors}}$

Consider this product $y=\frac{10^p-1}{9} \times \sum_{i=0}^{q-1} 10^{ip}$ , where $p,q \geq 2$ are integers.

On evaluation we get, $y=\frac{10^{pq}-1}{9}$ .

Both the numbers in the product are integers. The former being a string of 1's of length $p$ and the latter is a string of length $(q-1)p +1$ with every $p$ -th number being '1' and the rest being 0.

As it is assumed that $p,q \geq 2$ , both numbers are greater than 1 and hence $y$ is composite.

Since, $2015 = 5\times403=13\times155=31\times65$ , $x$ has at least 6 factors which are proper (two from each combinations of $p,q$ taken from the above factorizations).