Show that if $p$ is a prime number different from 2 and 5, then it divides at least one of the elements of the set $\left \{ 1,11,111,1111,...\right \}$.


Expanding on my comment, here is a direct solution.

Consider the sequence $u_{n+1}=10u_n\mod p$, with $u_0=1$. It can undertake only a finite number of values, so it's periodic, with some period $k$.

You have also that $u_n=10^n\mod p$, so periodicity reads:


And since $p$ is coprime to $10$, you can factor out $10^n$, and


That is, $p$ divides $10^k-1=999\dots999$.

If the prime number $p$ is greater that $5$, you are done, by dividing by $9$. The remaining case $p=3$ divides $111$.

Notice that repetitively computing $10u_n \mod p$, with $u_0=1$, is just another way to write the long division of $1$ by $p$.


Let $x=0.\overline{a_1a_2a_3\dots a_n}$ be the decimal number with $n$ periodic decimals in base $10$. Then: $$10^nx=\overline{a_1a_2a_3\dots a_n}.\overline{a_1a_2a_3\dots a_n}$$ $$10^nx-x=\overline{a_1a_2a_3\dots a_n}$$ $$(10^n-1)x=\overline{a_1a_2a_3\dots a_n}$$ $$x=\dfrac{\overline{a_1a_2a_3\dots a_n}}{10^n-1}$$ From this you can see that any number $\overline{a_1a_2a_3\dots a_n}$ is divisible by $10^n-1$ iff $x$ have periodic decimals. There is no positive integer which divided by $2$ and $5$ give periodic decimal number, so for any other number it is divisible by $10^n-1$.


The transformation $f(x) = 10x + 1$ is invertible in $\mathbb{Z}_p$ for $p$ coprime with 10. Since $f(0) = 1$, 0 and 1 are on the same orbit. The set $\{1, 11, 111, \ldots\}$ is precisely this orbit.


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