elementary number theory - Solve the congruence relation $x^nequiv 2pmod {13}$.

Consider the congruence $x^n\equiv 2\pmod {13}$. This congruence has a solution for $x$ if

(A) $n=5$.

(B) $n=6$.

(C) $n=7$.

(D) $n=8$.

I apply Chinese remainder theorem to solve it but I am fail. Can anyone help me please ?

Update :(18th Nov)

In the given answer I am unable to understand the step $2^A\equiv 1 \pmod{13}$ implies $12$ divides $A$. It's justification in comment is computational. I want an analytical answer.

Answer

Result: Let $p$ be a prime and $(a,p)=1$. Then the congruence $x^k\equiv a(\text{mod} ~p)$ has a solution iff $a^{(p-1)/d}\equiv 1(\text{mod}~p)$ where $d=(k,p-1)$

In your question $p=13$, $d=1,2,3,4,6$ Check for which $d$ the congruence

$$2^{12/d}\equiv 1(\text{mod}~ 13)$$ has a solution.

You'll get $d=1$ so $(k,12)=1$ implies $k=5,7,11,...$