Prove that number made only of $91$ ones is composite

Question

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.

Answer 1

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.

Answer 2

Note that

$$111\cdots 1 = \frac{999\cdots 9}{9} = \frac{10^{91}-1}{9} =\frac{(10^{7})^{13}-1}{9} =\frac{ (10^{7}-1)((10^{7})^{12} + (10^{7})^{11} + \cdots + 1)}{9} $$

$$ = \frac{ 10^{7}-1}{9} ((10^{7})^{12} + (10^{7})^{11} + \cdots + 1) =1111111\cdot((10^{7})^{12} + (10^{7})^{11} + \cdots + 1)$$

Answer 3

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 $u_{13}$.

Answer 4

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.