Prove that number made only of $91$ ones is composite How to prove that number  $11...1$ (there are exactly $91$ ones) is composite? I considered this number modulo some small primes, but none of them worked.
 A: Notice that $91=13\cdot 7$. Consider the number $n=\sum_{k=0}^{12}10^k$, i.e. the number $111\cdots1$ with $13$ ones. Then
$$\sum_{k=0}^{90}10^{k}=n+n\cdot 10^{13}+n\cdot 10^{26}+\cdots+n\cdot 10^{78}.$$
In general you have the following
$$\left(\sum_{k=0}^{p-1}10^k\right)\cdot\left(\sum_{h=0}^{q-1}10^{hp}\right)=\sum_{h=0}^{q-1}\sum_{k=0}^{p-1}10^{k+hp}=\sum_{n=0}^{pq-1}10^n.$$
Thus any number made of a composite number of ones is still composite.
A: Note that 
$$111\cdots 1 = \frac{999\cdots 9}{9} = \frac{10^{91}-1}{9} =\frac{(10^{7})^{13}-1}{9} =\frac{ (10^{7}-1)((10^{7})^{12} + (10^{7})^{11} + \cdots + 1)}{9} $$
$$ = \frac{ 10^{7}-1}{9} ((10^{7})^{12} + (10^{7})^{11} + \cdots + 1) =1111111\cdot((10^{7})^{12} + (10^{7})^{11} + \cdots + 1)$$
A: Since
$u_n=111...111$ with $n$ ones
is
$\frac19(10^n-1)$,
if
$n = ab$,
and using the factorization
$x^m-1
=(x-1)\sum_{j=0}^{m-1} x^j
$,
$\begin{array}\\
u_n
&=\frac19(10^n-1)\\
&=\frac19(10^{ab}-1)\\
&=\frac19(10^{a}-1)\sum_{j=0}^{b-1}10^{aj}\\
\end{array}
$
so $u_n$
is divisible by
$u_a$
(and also by $u_b$).
Therefore,
as Ender Wiggins pointed out,
since
$91=7\cdot 13$,
$u_{91}$
is divisible by
$u_7$ and $u_{13}$.
A: Try generalizing the following:
$$
111111=1001\cdot111=10101\cdot11.
$$
Added: The more general phenomenon is that numbers whose base-$10$ representation has a repeating pattern, where the repeating unit consists of more than a single digit, are necessarily composite.  For example,
$$
762127621276212=76212\cdot10000100001.
$$
Our example $111111$ fits this description, since it can be thought of as two repetitions of the unit $111$ or as three repetitions of the unit $11$.
Interestingly, if the repeating pattern is only one digit long, as in $77777$, the number is, again, composite, as long as the digit is not $1$:
$$
77777=7\cdot11111.
$$
The only case left is the case of repeating $1$s with a prime number of digits.  These may very well be prime.  Apart from $11$, the smallest example is the number consisting of $19$ $1$s.
