How to prove that $11^{n+2} + 12^{2n+1}$ can be divided by $133$ where $n\in\mathbb N$

Question

I know that I have to get to a point where an summation is multiplied by $133$ but I don't know how to get there. I tried formating the equation like $11^2\times11^n + 12 \times 12^ {2n}$ but I got stuck there and I don't really know how to prove it with induction. I really need to solve this.

Thank you for your help and sorry if I misspelled anything. Also sorry that you have to edit it, I don't really know how to make superscripts.

The question will probably be closed unless you edit the question and show some effort and explain what you've tried and where you are stuck. — Nick, Dec 12 '20 at 20:35
I don't really understand induction so i tried formating the equation like 111111^n+12*12^n and solve it that way but i am stuck. But if you know how, could you tell me how to do that? — WAT, Dec 12 '20 at 20:35
If $n=1$, we get $11^3+12^2=1475$ which is not even divisible by $7$. — lulu, Dec 12 '20 at 20:36
If so, then I'll try to give an answer before the q is closed (unless you change it). — A-Level Student, Dec 12 '20 at 20:39
The correct statement is this duplicate. So no need to prove this again. — Dietrich Burde, Dec 12 '20 at 20:43
Or this : https://math.stackexchange.com/questions/716789/proving-divisibility-by-using-induction-133-mid-11n2-122n1?noredirect=1&lq=1 — A-Level Student, Dec 12 '20 at 20:45
Yes, sorry for everyone, i didn't read it right. I fixed it. And also sorry for the bad superscript. — WAT, Dec 12 '20 at 20:48
@WAT that's all right :) Here's a great guide to MathJax (what we use to fornat equations etc): https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — A-Level Student, Dec 12 '20 at 20:50

A-Level Student · Accepted Answer · 2020-12-12T21:47:00.007

1

Seeing as you don't seem to be familiar with induction, I'll give a quick explanation.

We begin by checking the base case for the statement, ie for the first value we are interested in- in your case, check for when $n=1$. Next, we assume the statement is true for some $n=k$.

Next, we use our assumption to show that if the assumption is true then the statement is also true for $n=k+1$. But we already know the statement is true for $n=1$ (in your case), so it must be true for $n=2,3,4,...$ and so on.

I can provide a proof by induction in this case if you feel you are ready for it?

Proof by induction I will make this proof quite explicit as this is your first proof by induction. First, let $f(n)=11^{n+2}+12^{2n+1}$.

Base case

Check for $f(1)$: $$11^3+12^3=133\times23$$ which is certainly divisible by $133$!

Assumption

Assume true for $n=k$: $$f(k)=11^{k+2}+12^{2k+1}\text{is divisible by $133$ for all $k\in\mathbb N$}$$

Inductive step

Now to prove that the assumption implies the truth of the statement for $n=k+1$:

$$f(k+1)=11^{k+3}+12^{2(k+1)+1}=11^{k+3}+12^{2k+3}=11(11^{k+2})+144(12^{2k+1})$$ $$=11(11^{k+2}+12^{2k+1})+133(12^{2k+1})=11f(k)+133(12^{2k+1})$$ Now, $133(12^{2k+1})$ is certainly divisible by $133$, but by the assumption, $f(k)$ is also divisible by $133$. Hence we have proved that $f(k+1)$ is divisible by $133$ if $f(k)$ is divisible by $133$. But by the base case, we know it to be true for $n=1$. Hence the result follows for all $n\in\mathbb N$. I'll just formally write the conclusion below.

Conclusion

If $f(n)$ is divisible by $133$ for $n=k$ and this implies $f(k+1)$ is divisible by $133$, as $f(1)$ is divisible by $133$ then $f(n)$ is divisible by $133$ for all $n\in\mathbb N$ by the principle of mathematical induction.

edited Dec 12 '20 at 21:47

answered Dec 12 '20 at 20:56

A-Level Student

6,520

Yes I think i got it. If you can provide a proof I would really appreciate it. Thank you. – WAT Dec 12 '20 at 21:11
@WAT please see the added proof. – A-Level Student Dec 12 '20 at 21:22
Thank you very much. You are brilliant. – WAT Dec 12 '20 at 21:26
@WAT you are most welcome, but I don't think I'm brilliant, rather I've simply been taught this method :)) You can accept this answer if it was very helpful ;) – A-Level Student Dec 12 '20 at 21:27
Could you help me with some other discrete mathematics problems? – WAT Dec 12 '20 at 22:03
@WAT I could have a go at them. I'm only in high school myself. – A-Level Student Dec 12 '20 at 22:21
Is there a way i can reach you? Because I can't post any more questions for 3 days – WAT Dec 12 '20 at 22:27
I'd prefer not to give any personal details out, sorry. Maybe I'll make an email adress just for maths, I'll tell you if I do. – A-Level Student Dec 12 '20 at 22:29
Can you give a try to this question: https://math.stackexchange.com/questions/3946229/if-n100-mod-73-57-and-n101-mod-73-11-then-what-is-n-mod-73-where – WAT Dec 12 '20 at 22:32
@WAT you can email [email protected] for any help with other q's. – A-Level Student Dec 12 '20 at 22:34
I have sent an email – WAT Dec 12 '20 at 22:46

score 1 · Answer 2 · answered Dec 13 '20 at 02:45

1

If you know modular arithmetic, it's easy:

$$11^{n+2}+12^{2n+1}=121\cdot11^{n}+12\cdot144^{n}\equiv121\cdot11^{n}+12\cdot11^{n}=133\cdot 11^{n}\equiv0\bmod133.$$

answered Dec 13 '20 at 02:45

J. W. Tanner

60,406

How to prove that $11^{n+2} + 12^{2n+1}$ can be divided by $133$ where $n\in\mathbb N$

2 Answers2