If $11..11$($n$ times 1) is prime, prove that $n$ is also prime

Question

We know 11...11 (n times) is a prime number prove n is also prime. it is easy to understand n is an odd number and 3 does not divide it but I could not do some thing more

Answer 1

$$ \mathrm A=\underbrace{111\ldots111}_{n \text{ digits}}=\frac{10^n-1}{9}$$

Now, let us assume $n$ isn't a prime, but $\mathrm A$ is. That is, $n =a\cdot b$ for some integers $a$ and $b$, greater than $1$.

\begin{align} \mathrm A&=\frac{10^{ab}-1}{9}\\ &=\frac{(10^a)^b-1}{9}\\ &=\frac{(10^a-1)}{9}(1+(10^a)+(10^a)^2+\ldots (10^{a})^{b-1})\\ &=(1+10+10^2+\ldots+10^{a-1})(1+(10^a)+(10^a)^2+\ldots (10^{a})^{b-1})\\ \end{align}

Thus we have a factor of $\mathrm A$, that is $\frac{10^a-1}{9}$ which is an integer greater than $1$ (Since $a>1$), giving us a contradiction that $n$ is composite.

Hence, $n$ is a prime number.

Answer 2

$$ \underbrace{111111}_{6\text{ digits}} = 11\cdot 10101 $$ $$ \underbrace{111111111111}_{12\text{ digits}} = 1111\cdot 100010001 $$ and in a similar way if a repunit is made by a composite number of digits, it cannot be prime.

Answer 3

We could write your number as $11\sum_\limits{i=0}^{n-1} 100^i$

which is a finite geometric series:

$11\sum_\limits{i=0}^{n-1} 100^i = 11\frac {100^n - 1}{99} = \frac 19 (100^n - 1)$

If $n$ is composite, i.e. $n=pq,$

$(100^{pq} -1)$ has $(100^p-1)$ and $(100^q-1)$ as factors (among others).

Answer 4

Define $R_n = \sum_0^{n-1}10^i$, the number consisting of $n$ occurrences of the the digit $1$.

Then if $n=pq$, with $p,q>1$, we have that $R_p \mid R_n$ since $\sum_0^{n-1}10^i = \left(\sum_{i=0}^{p-1}10^i\right)\left(\sum_{j=0}^{q-1}10^{jp}\right)$ (and similarly $R_q \mid R_n$). For example, $R_{15} = 10000100001 \cdot R_{5}$

Answer 5

The trick to notice is $1111.....1$ with $k$ ones times $1000....010........01$ with $j$ ones and $k-1$ zeros between them will result in $11111.....111$ with $j*k$ ones. (The $k$ ones will "line up" perfectly with the $1$ followed by $k - 1$ zeros and the $j$ repetitions will give us $j$ reptitions of the $k$ ones.)

That's not a very formal proof but:

If $n = j*k$ then $1111.....$ with $n$ ones is $\sum\limits_{i=0}^{n-1} 10^i$.

And $(\sum\limits_{i=0}^j 10^k)*(\sum\limits_{i=0}^{k-1} 10^i)$

$=\sum\limits_{m=0;m+k}^{n} \sum\limits_{l = 0}^{k-1}10^{m*k + l}$

$=\sum\limits_{i=0}^{jk-1=n-1}10^i$

So $1111.....$ with $n$ is not prime if $n$ is not prime.

Is a formal proof.