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?
On
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 $1321$.
On
The prime factorization of the number is:
$$41 \cdot 271 \cdot 1321 \cdot 21649 \cdot 62921 \cdot 513239 \cdot 83251631 \cdot 1300635692678058358830121$$
obtained by FactorInteger[] in Mathematica.
On
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.
On
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.
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 $$x=110000 + 1100 + 11 = 10101\cdot11.$$
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.
On
You can either do five blocks of eleven 1s:
$$11111111111\ 11111111111\ 11111111111\ 11111111111\ 11111111111 \\ =11111111111×100000000001000000000010000000000100000000001$$
or you can do eleven blocks of five 1s:
$$11111\ 11111\ 11111\ 11111\ 11111\ 11111\ 11111\ 11111\ 11111\ 11111\ 11111 \\ =11111×100001000010000100001000010000100001000010000100001$$
It turns out, in this case, that all these factors are themselves composite.
(Answer taken from comment by Jeppe Stig Nielsen.)
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).