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.
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Sign up to join this communityProblem: 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).