Is $1111111111111111111111111111111111111111111111111111111$ ($55$ $1$'s) a composite number? This is an exercise from a sequence and series book that I am solving.
I tried manipulating the number to make it easier to work with:
$$111...1 = \frac{1}9(999...) = \frac{1}9(10^{55} - 1)$$
as the number of $1$'s is $55$.
The exercise was under Geometric Progression and Geometric Mean. However, I am unable to think of a way to solve this problem using GP.
How do I proceed from here?

This number can be written as $$x=\sum_{k=1}^{55}10^{k-1}$$ i.e. it has the following form: $$x=\sum_{k=1}^{n}a^{k-1}$$
Such numbers are always composite if $n$ is composite. Let $n = pq$, where $p,q\in\mathbb N$. Then,
$$x=\sum_{k=1}^{pq}a^{k-1} = \sum_{l=1}^p \sum_{m=1}^q a^{q(l-1)+(m-1)}=
\left(\sum_{l=1}^p a^{q(l-1)}\right)\left(\sum_{m=1}^q a^{(m-1)}\right)$$
Since, $55$ is composite, it follows that $\sum_{k=1}^{55}10^{k-1}$ is composite as well.
Simple illustrative example
To get the idea behind this construction, consider a simpler example of $$x = 111111=\sum_{k=1}^{6}10^{k-1}.$$ This number can be factorized using this construction (and the fact that $6=2\cdot3$) as $$x = 111000 + 111= 1001\cdot111$$ or $$x=110000 + 1100 + 11 = 10101\cdot11.$$
General conclusion
If a natural number $x$ can be represented with a string of composite number of repeating $1$s (regardless of the base of the numeral system), then $x$ is composite.
