3
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ If I may ask though, what is the number of factors for $2^{2019}+1$? $\endgroup$ – Mike Apr 29 at 20:47
  • 1
    $\begingroup$ @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. $\endgroup$ – Winther Apr 29 at 21:07
  • $\begingroup$ Thanks @Winther. Which makes me wonder if there is an elegant trick for solving this. The exponent 2019 i.e., the current year makes me think this is doubles as a contest math problem, which have some sort of snappy solution. $\endgroup$ – Mike Apr 29 at 21:44
  • $\begingroup$ Oh, just show that infinitely many are divisible....Then yes it is easy to see from the above. For whatever reason I misread $\endgroup$ – Mike Apr 29 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.