# Proving $(11^n -6)$ is divisible by $5$. [duplicate]

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.

• $11^{n+1}-6$ "...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 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 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 at 16:59
• Perfect. I will think on that and give it a go. Thank you for the pointer! – Goldsten Jan 14 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 at 17:02

Hint:

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

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

• 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 at 16:56
• or polynomial remainder theorem @Goldsten – user645636 Jan 14 at 20:01
• Interesting suggestion. I will have to look into that further. Thank you for the suggestion @RoddyMacPhee! – 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 $$5$$

• Thank you so much. I knew it had to be something small I was missing. – 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.

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

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

• 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 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... – user645636 Jan 14 at 19:59
• @RoddyMacPhee That's cute too. (That's actually really cute). – fleablood Jan 15 at 1:36
• @fleablood and other than the concept of induction and proving the base case it's grade school level. – user645636 Jan 15 at 1:55