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!

$$P_1(x)=\frac{x^{24}-1}{x^2-1}$$
$$P_2(x)=\frac{x^{12}-1}{x-1}$$
$$\frac{P_1(x)}{P_2(x)}=\frac{x^{24}-1}{x^{12}-1}\cdot\frac{x-1}{x^2-1}=\frac{x^{12}+1}{x+1}$$
Then Ruffini's rule tells us that the remainder of this reduced division is the polynomial $x^{12}+1$ evaluated at $-1$, i.e. 2. When the top and bottom of $\frac2{x+1}$ are multiplied by $\frac{x^{12}-1}{x^2-1}$, the denominator becomes $P_2(x)$ and the numerator gives the final answer of $\frac{2(x^{12}-1)}{x^2-1}=2+2x^2+2x^4+2x^6+2x^8+2x^{10}$.
