Find $k$: $27^k\equiv 2\mod 2021$

Question

By using these: $$2^{11}=2048\equiv 27\mod 2021$$ $$2021=43\cdot 47$$

I should find integer $k$

$$27^k\equiv 2\mod2021$$

Well, I can only come up with something like this: $$a^{42}\equiv 1\mod 43$$ $$b^{46}\equiv 1\mod 47$$

$$\ a\ne 0 \mod 43{,}\quad b\ne 0 \mod 47$$

Answer 1

Since you know that $2^{11}\equiv 27 \pmod{2021}$, suppose that $27^k\equiv 2 \pmod{2021}$. Then $(2^{11})^k\equiv 2^{11k}\equiv 2\equiv 2^{1+j\phi(2021)} \pmod{2021}$. Can you find $k$ and $j$ with $11k=1+j\phi(2021)$?