Tell whether $\dfrac{10^{91}-1}{9}$ is prime or not?

Question

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.

Answer 1

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.

Answer 2

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 $n$.

Answer 3

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 $\displaystyle\frac{a^{91}-1}{a-1}$

Answer 4

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.