0

Show that every positive integer relatively prime to $10$ divides infinitely many repunits. (Hint: Note that the $n-$digit repunit $111\cdots1 = \dfrac{(10^{n}-1)}{9}$.)

Let $(a,10)=1$ and $n=9k{\phi(a)}$ using eulerphi function, where $k$ is positive integer.

When $(a,9)= 1 , 3$ it is okay because $81$ and $a$ divides $10^{n}-1$ by Binomial theorem and CRT.

So $a$ divides $\dfrac{(10^{n}-1)}{9}$.

But I don t know when $(a,n)=9$, Must $a$ divides $\dfrac{(10^{n}-1)}{9}$ ? I dont think so

Then how can i solve it?

user14111
  • 3,118
wtgrea
  • 109
  • Same as in the dupe, except use Euler's totient $\phi$ instead of little Fermat (or, more generally, use that permutation orbits are cycles as I explain there). $\ \ $ – Bill Dubuque May 05 '25 at 04:05

1 Answers1

0

Let $a$ and $10$ be relatively prime. If $9a$ divides $10^n-1$, then $a$ will divide $\frac{10^n-1}{9}$.

So let $n=k\varphi(9a)$.

André Nicolas
  • 514,336