# Infinitely many of $1,11,111,\dots$ is divisible by $2^{2019}+1$

Problem: Show that infinitely many members of the sequence $$1,11,111,1111,11111,11111, \dots$$ are divisible by $$2^{2019}+1$$.

I solved this problem using Euler's totient theorem. What I'm wondering is what, if any, other ways are there.

Your sequence is $$(10^n-1)/9$$, so what you need is $$10^n \equiv 1 \bmod 9 (2^{2019}+1)$$. After verifying that $$\gcd(10, 9(2^{2019}+1))=1$$, this follows from the fact that the multiplicative group of integers modulo $$9(2^{2019}+1)$$ is a finite group. (Euler's totient theorem gives the order of the group, but all you need is the fact that it is a finite group).
• If I may ask though, what is the number of factors for $2^{2019}+1$?
• @Mike It's easy to find some factors as it's one the form $2^{ab}+1$ ($3\cdot 673=2019$) and $x^{ab} + 1 = (x^a+1)(1-x^a + x^{2a}-\ldots+x^{a(b-1)})$. It therefore has atleast $5$ factors, but finding them all (which we need to know how many there are) requires factoring a number with $400$ digits, which is not that easy. Apr 29, 2019 at 21:07