Calculation of $2^{XXXX} + 3^{XXXX}\pmod{11}$

Question

I've a question: How do I calculate $2^{2020} + 3^{2020}\pmod{11}$?

Is there a theorem or any trick to do it? I need to show all the steps I used to calculate the Rest but I've no clue how to even start with the calculation...

Does anyone have any tips and could show me how to calculate it?

Greetings!!!

score 3 · Answer 1 · answered Oct 30 '19 at 20:48

3

Hints:

It suffices to compute $2^{2020} \mod 11$ and $3^{2020} \mod 11$ separately, and add them together modulo $11$.
Try computing $2^n \mod 11$ for $n = 1, 2, 3, \ldots$. Do you see a pattern?
Do the same for $3^n \mod 11$.

answered Oct 30 '19 at 20:48

angryavian

89,882

Thank you very much for your answer!!! I could solve it now by lookin at them separately as you adviced me to! – Default Nov 07 '19 at 15:37

score 3 · Answer 2 · answered Oct 30 '19 at 20:50

3

Hint:

By Fermat's little theorem, $2^{10}\equiv3^{10}\equiv1\mod 11$.

answered Oct 30 '19 at 20:50

J. W. Tanner

60,406

score 1 · Answer 3 · answered Oct 30 '19 at 20:51

1

The answer of angryavian is great. As a small addendum: Once $\mathbb{Z}_{11}$ is a field, because $11$ is prime, so there are $n,m$ naturals s.t. $2^n,3^m$ are both $\equiv 1\pmod{11}$. You could try to find these numbers. It would be very good to computation, right?

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Edit: The answer above is the great Theorem for that, I was typing before the answer. Well, hope all clues can help you. Good studies!

answered Oct 30 '19 at 20:51

Quiet_waters

1,493

1

Thank you very much for your help!!! I could solve it now!!! But it took me really a long time to see the pattern and then come to the result... – Default Nov 07 '19 at 15:39
You're welcome, Default. – Quiet_waters Nov 07 '19 at 17:40

Calculation of $2^{XXXX} + 3^{XXXX}\pmod{11}$

3 Answers3