Show that infinitely many members of $1,11,111,1111,\dots$ are divisible by $2^{2017}+1$.

Question

Show that infinitely many members of $1,11,111,1111,\dots$ are divisible by $2^{2017}+1$.

I know that the sequence above is equal to $a_n = \frac{10^n-1}{9}$ for $n = 1,2,3,\dots$. I also feel that I should use the pigeonhole principle but do not know how to attack the problem.

This seems like a recent competition problem. Where is it from? — Arthur, Nov 06 '17 at 15:26
@Arthur I've seen problems like this in competitions but it was actually just a problem given by my instructor — user499939, Nov 06 '17 at 15:39

score 1 · Answer 1 · answered Nov 06 '17 at 15:33

1

Hint: If $10^n-1$ is divisible by $m$, then so is $10^{kn}-1$ for all $k \in \mathbb N$.

answered Nov 06 '17 at 15:33

lhf

216,483

Show that infinitely many members of $1,11,111,1111,\dots$ are divisible by $2^{2017}+1$.

1 Answers1