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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.

Bill Dubuque
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Goldsten
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    $11^{n+1}-6$ "...my first thought was to factor 11 out..." Good., now remember that $11=10+1$ and continue as $=11\cdot 11^n - 6 = (10+1)\cdot 11^n - 6 = 5\cdot 2\cdot 11^n + (11^n-6)$ – JMoravitz Jan 14 '20 at 16:50
  • Brilliant! That is a fantastic idea. This is much easier to prove with that hint given. Thanks a million! – Goldsten Jan 14 '20 at 16:54
  • you are also trying right just use $n-1$ insted $n$ and $n$ insted of $n+1$ you will get your answer – TheStudent Jan 14 '20 at 16:59
  • Perfect. I will think on that and give it a go. Thank you for the pointer! – Goldsten Jan 14 '20 at 17:00
  • 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 $5$. – fleablood Jan 14 '20 at 17:02
  • ... but working with $11^{n+1} - 6$ we can get $11^{n+1} -6=11^{n+1} -1 -5 = (11-1)(1+ 11+ 11^2 + ..... + 11^n) - 5 = 10(1+11+11^2 + ... + 11^n) -5=5[2(1+11+ .... + 11^n)- 1]$. But that isn't induction. – fleablood Jan 14 '20 at 17:04
  • I see. Thank you so much for pointing that out to me. I seemed to have overlooked that detail. Hmmmm. Do you have any recommendations on how I might approach this inductively? – Goldsten Jan 14 '20 at 17:18
  • Never mind, I just noticed the solution by @JMoravitz. Thank you all so much! – Goldsten Jan 14 '20 at 17:25

3 Answers3

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Hint:

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

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

cansomeonehelpmeout
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  • Fantastic approach. Thank you so much. I think I will do just that and utilize the binomial theorem to finish the proof. – Goldsten Jan 14 '20 at 16:56
  • or polynomial remainder theorem @Goldsten –  Jan 14 '20 at 20:01
  • Interesting suggestion. I will have to look into that further. Thank you for the suggestion @RoddyMacPhee! – Goldsten Jan 14 '20 at 20:12
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$11=1 mod $ $5$ and $6=1$ $mod 5$ implies that $11^n=1$ mod $5$ and $11^n-6=0$ mod $5$

Tsemo Aristide
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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.

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

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

fleablood
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  • 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! – Goldsten Jan 14 '20 at 16:58
  • 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... –  Jan 14 '20 at 19:59
  • @RoddyMacPhee That's cute too. (That's actually really cute). – fleablood Jan 15 '20 at 01:36
  • @fleablood and other than the concept of induction and proving the base case it's grade school level. –  Jan 15 '20 at 01:55