What is the remainder of $2019^{2021}$ divided by 11

Question

With these kinds of problems, I use congruence, and I aim to get the remainder to be -1, which then gets me pretty close to the solution. On this one I'm kinda stuck. Can't get to that -1.

Answer 1

A very elementary way to calculate this is the following:

First, divide $2019$ by $11$ to get $2019=183\cdot11+6$. Note that $$(183\cdot11+6)^{2021}=6^{2021}+\text{ a multiple of }11$$ so we can calculate the remainder of $6^{2021}/11$ instead.

Now, if you try with lower exponents, can find this:

• For $6^1$ the remainder is $6$
• For $6^2$ the remainder is $3$, that is, $6^2=11k_2+3$
• For $6^3$ we can do $6^3=6(11k_2+3)=11k_3+7$
• For $6^4$ we can do $6^4=6(11k_3+7)=11k_4+9$

...

• For $6^{11}$ the remainder is $6$ again.

This suggests that the sequence of the remainders is periodic, and the period is $10$. (Note that we can find the remainder of a power knowing only the remainder of the previous power). That is, if you increase the exponent by $10$ the remainder is the same.

So $6^{2021}/11$ has the same remainder as $6^1/11$, that is, $6$.

Remark: Fermat's theorem, mentioned in other answers, allows to find the period much faster.

Answer 2

We recall Fermat's little theorem for any prime, p, and $a\ncong 0$ mod p $$a^{p-1}\cong1\text{mod}p$$ $2019\cong6$mod $11$ and $2021\cong1$ mod 10 thus a little rearranging of your equation gives $2019^{2021}\cong 6^{10\gamma+1}\cong 6^{10^\gamma}6\cong6$

Answer 3

The remainder of $2019\bmod 11$ is, according to a well known rule, equal to $9-1+0-2=6$. To have the remainder of $2019^{20121}\equiv 6^{2021}$, apply lil' Fermat.