Is $1111111111111111111111111111111111111111111111111111111$ ($55$ $1$'s) a composite number?

You can see with some very quick "in your head" maths that this is a composite number. Since 55 is 5 times 11 we can take a number that is 11111111111 (eleven ones) and divide the original by it using standard long division. Its easy to see that the result will be 100000000001000000000010000000000100000000001.

Similarly you can see trivially see that 11111 is a factor of the number.

The number is composite.

We have

\begin{align*} \underbrace{11\ldots111}_{55 \text{ times }} = \frac{1}{9} \cdot (10^{55} - 1) \\ = \frac{1}{9} \cdot ((10^{5})^{11} - 1) \\ \end{align*}

Also, note that $x^{m} - 1$ is divisible by $x - 1$. Here, we can plug in $x = 10^{5}$ and $m = 11$. As a result, we see that the quantity is divisible by $99999$, meaning that the number must be divisible by $11111$ (and hence, composite).

More explicitly, $$\begin{align*} \frac{10^{55} - 1}{9} &= \frac{(10^5)^{11} - 1}{9} \\ &= \frac{(10^5 - 1)(10^{50} + 10^{45} + 10^{40} + \cdots + 10 + 1)}{9} \\ &= \frac{(10 - 1)(10^4 + 10^3 + 10^2 + 10^1 + 1)(10^{50} + 10^{45} + \cdots + 1)}{9} \\ &= (10^4 + 10^3 + \cdots + 1)(10^{50} + 10^{45} + \cdots + 1). \end{align*}$$ The first factor is $11111$, which in turn is $41 \cdot 271$, and the second factor has as its smallest prime factor $1321$.