I approached this inductively. For n=1, $11^1-6=5$, which is divisible by 5.

For n+1, 11^(n+1) -6, there must be some trick with the factorization that I am missing. My first thought was to factor 11 out, but that ultimately lead nowhere.

Thank you all for your time and assistance.

  • 4
    $\begingroup$ $11^{n+1}-6$ "...my first thought was to factor 11 out..." Good., now remember that $11=10+1$ and continue as $=11\cdot 11^n - 6 = (10+1)\cdot 11^n - 6 = 5\cdot 2\cdot 11^n + (11^n-6)$ $\endgroup$ – JMoravitz Jan 14 at 16:50
  • $\begingroup$ Brilliant! That is a fantastic idea. This is much easier to prove with that hint given. Thanks a million! $\endgroup$ – Goldsten Jan 14 at 16:54
  • $\begingroup$ you are also trying right just use $n-1$ insted $n$ and $n$ insted of $n+1$ you will get your answer $\endgroup$ – TheStudent Jan 14 at 16:59
  • $\begingroup$ Perfect. I will think on that and give it a go. Thank you for the pointer! $\endgroup$ – Goldsten Jan 14 at 17:00
  • $\begingroup$ With induction you need to relate $P(n+1)$ to $P(n)$. so you should relate $11^{n+1} - 6$ to $11^n -6$. Just working and $11^{n+1}-6$ by itself won't take advantage of you knowing that $11^n-6$ is divisible by $5$. $\endgroup$ – fleablood Jan 14 at 17:02


Since $11^n-6=11^n-1-5$, we have $$5\mid 11^n-6\iff 5\mid 11^n-1$$

Now write $$11^n=(10+1)^n$$

  • $\begingroup$ Fantastic approach. Thank you so much. I think I will do just that and utilize the binomial theorem to finish the proof. $\endgroup$ – Goldsten Jan 14 at 16:56
  • $\begingroup$ or polynomial remainder theorem @Goldsten $\endgroup$ – user645636 Jan 14 at 20:01
  • $\begingroup$ Interesting suggestion. I will have to look into that further. Thank you for the suggestion @RoddyMacPhee! $\endgroup$ – Goldsten Jan 14 at 20:12

$11=1 mod $ $5$ and $6=1$ $mod 5$ implies that $11^n=1$ mod $5$ and $11^n-6=0$ mod $5$

  • $\begingroup$ Thank you so much. I knew it had to be something small I was missing. $\endgroup$ – Goldsten Jan 14 at 16:53

If you are going to do it inductively:

As $11^n -6$ is divisible by $5$ you have to show

$11^{n+1} - 6 = (11^n-6) + 5k$ or in other words that

$5$ divides $(11^{n+1} - 11^n)$ and I'm sure you can factor that.

$11^{n+1} - 11^n = 11^n(11 -1) = 11^n*10 = 5*(11^n*2)$.


But you don't have to do it inductively. $11\equiv 1 \pmod 5$ so $11^n\equiv 1\pmod 5$.

And $6\equiv 1 \pmod 5$

So $11^n - 6 \equiv 1-1\equiv 0\pmod 5$.

  • $\begingroup$ Thank you so much. The modular approach is the more familiar approach for me, and one that I (Unfortunately) did not notice at first. Thank you so much for the assistance! $\endgroup$ – Goldsten Jan 14 at 16:58
  • $\begingroup$ or you can multiply the first by 11 getting $11^{n+1}-66=(11^{n+1}-6)-60$ and 60 divides by 5 and multiplying doesn't get rid of factors so $11^{n+1}-66$ is divisible by 5, therefore, so is our next number... $\endgroup$ – user645636 Jan 14 at 19:59
  • $\begingroup$ @RoddyMacPhee That's cute too. (That's actually really cute). $\endgroup$ – fleablood Jan 15 at 1:36
  • $\begingroup$ @fleablood and other than the concept of induction and proving the base case it's grade school level. $\endgroup$ – user645636 Jan 15 at 1:55

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