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!!!

Answer 1

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 $3^n \mod 11$.

Answer 2

Hint:

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

Answer 3

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!