Question: I have the following equations :

$$x^5=2 \pmod {13}$$ $$x^3=2 \pmod{11}$$

I wonder how to solve such equations is there a method to get rid off the powers in order to use the Chinese Remainder Theorem without checking each $x$ value?

Any ideas?

Thank you.

Answer: 
**Hint:** If $x^3\equiv 2\pmod{11}$ then $ x^{21}\equiv 2^7\pmod{11}$.

By Fermat, we have $x^{21}\equiv x\pmod{11}$.
we have $$1^5\equiv 1 \mod 13$$
$$2^5\equiv 6 \mod 13$$
$$3^5\equiv 9 \mod 13$$
$$4^5\equiv 10 \mod 13$$
$$5^5 \equiv 5 \mod 13$$
$$6^5 \equiv 2 \mod 13$$
can you proceed?