Remainder when the polynomial $1+x^2+x^4+\cdots +x^{22}$ is divided by $1+x+x^2\cdots+ x^{11}$ 
Question : Find the remainder when the polynomial $1+x^2+x^4+\ldots +x^{22}$ is divided by $1+x+x^2+\cdots+ x^{11}$.

I tried using Euclid's division lemma, I.e.
$$P_1(x)=1+x^2+x^4+\cdots+x^{22}$$
$$P_2(x)=1+x+x^2+\cdots+x^{11}$$
Then for some polynomial $Q(x)$ and $R(x)$; we have 
$$P_1(x)=Q(x)\cdot P_2(x)+R(x)$$
Now, we put the values of $x$ such that $R(x)=0$ and form equations, but this method is way too long and solving the 11 set of equations for 11 variable (Since $R(x)$ a polynomial of at most 10 degree) is impossible to do for a competitive exam where the average time for solving a question is 3 minutes.
Another method is using the original long division method, and following the pattern, we can predict $Q(x)$ and $R(x)$, but it's also very hard and time taking.
I am searching for a simple solution to this problem since last a week and now I doubt even we have a simple solution to this question.
Can you please give me a hint/solution on how to proceed to solve this problem in time?
Thanks!

\begin{eqnarray*}
p_2(x) &=& (x^6+1)(x^5+x^4+x^3+x^2+x+1)\\
       &=& (x^6+1)(x^3+1)(x^2+x+1)\\
       &=& (x^6+1)(x+1)(x^2-x+1)(x^2+x+1)\\
       &=& (x+1)(x^6+1)(x^4+x^2+1)
\end{eqnarray*}
\begin{eqnarray*}
p_1(x) &=& (x^{12}+1)(x^{10}+x^8+x^6+x^4+x^2+1)\\
       &=& (x^{12}+1)(x^6+1)(x^4+x^2+1)
\end{eqnarray*}
Write $$p_1(x) = k(x)p_2(x) +r(x)$$ then $(x^6+1)(x^4+x^2+1)$ divides $r(x)$. 
So $r(x) = (x^6+1)(x^4+x^2+1)s(x)$ where $s(x)$ is constant. So $$x^{12}+1 = k(x)(x+1)+ s(x)$$ Put $x=-1$ and we get: $s(-1)= 2$.
