How to prove 1111......11 (91 digits) is a prime or composite number?

Question

How to prove $1111......11$ ($91$ digits) is a prime or composite number?

My Approach:

$1111......11$ can be expressed as $10^{0}+10^{1}+10^{2}+...…..+10^{90}$

Using summation of a geometric progression formula,

$$10^{0}+10^{1}+10^{2}+...…..+10^{90} = 1\frac{(10^{91}-1)}{(10-1)}=\frac{(10^{91}-1)}{(9)}$$

I do not know how to proceed after this step. Kindly guide me how to solve this problem.

Answer 1

If $m=pq$ where $p>1,q>1$ are natural numbers

using Why is $a^n - b^n$ divisible by $a-b$?

$$\dfrac{a^m-1}{a-1}=\dfrac{a^{pq}-1}{a-1}=\dfrac{(a^p)^q-1}{a-1}$$ will be divisible by $\dfrac{a^p-1}{a-1}$ which will be $>1$ as $p>1$

Answer 2

Let me illustrate on a smaller repunit:

$111111=11\cdot10101$ and $111111=111\cdot1001$, because $6=3\cdot2$

Yours is the same problem, except $91=7\cdot13$.

Answer 3

Note that$$91=13(7)$$ so you may factor $$1111111$$ and your number is $$1111111(10^{12\times {7}} + 10^{11\times 7}+10^{10\times 7}+...1)$$

So it is composite.