Tell whether $\dfrac{10^{91}-1}{9}$ is prime or not? I really have no idea how to start. The only theorem considering prime numbers I know of is Fermat's little theorem and maybe its related with binomial theorem.
Any help will be appreciated.
 A: Just think through the actual number.
$10^{91}$ is a $1$ with $91$ $0$'s after it.
$10^{91}-1$ is therefore $91$ $9$'s in a row.
$\frac{10^{91}-1}{9}$ is therefore $91$ $1$'s in a row.
Due to the form of this number, $x$ $1$'s in a row will divide it, where $x$ is a divisor of $91$.
For example $1111111$ is a divisor, so is $1111111111111$.
Hence the number is not prime.
A: Hint: Note that $91=7\cdot 13$, so the numerator is divisible by $10^7-1$. 
If this is not obvious, it may be clearer if we let $x=10^7$. Then $10^{91}=x^{13}$, and it is a familiar fact that $x-1$ divides $x^n-1$ for any positive integer $n$. 
A: Using Why $a^n - b^n$ is divisible by $a-b$?,
$$a^{91}-1=(a^{13})^7-1^7$$ is divisible by $a^{13}-1$ and by similarly by $a^7-1$
Again, using Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$,
$$(a^7-1,a^{13}-1)=a^{(7,13)}-1=a-1$$
Hence, $\displaystyle\frac{a^{13}-1}{a-1}, \frac{a^7-1}{a-1}$ are co-prime primes and divisors of $\displaystyle\frac{a^{91}-1}{a-1}$
A: You have
$$111111....1=1+10+10^2.....+10^{90}=(10^{84}+10^{77}+10^70.....+10^7+1) (10^6+10^5+10^4+ \dots + 10^0);$$
this is product of two integers 2 or greater,  so it is not prime.
