jmc

From the formula for a geometric series, $$p(x)=1+x^2+x^4+x^6+\cdots +x^{22}=\frac{x^{24}-1}{x^2-1}.$$Likewise, $$q(x)=1+x+x^2+x^3+\cdots +x^{11}=\frac{x^{12}-1}{x-1}.$$At first, it may look like we can write $p(x)$ as a multiple of $q(x)$: $$\frac{x^{24}-1}{x^2-1}=\frac{x^{12}-1}{x-1}\cdot \frac{x^{12}+1}{x+1}.$$Unfortunately, $\frac{x^{12}+1}{x+1}$ is not a polynomial. A polynomial of the form $x^n+1$ is a multiple of $x+1$ only when $n$ is odd.

So, we can try to get close by considering $\frac{x^{11}+1}{x+1}.$ Let's also multiply this by $x,$ so that we get a polynomial of degree 12. Thus, $$\begin{array}{rl}\frac{x^{12}-1}{x-1}\cdot \frac{x(x^{11}+1)}{x+1}& =\frac{x^{12}-1}{x-1}\cdot \frac{x^{12}+x}{x+1}\\ & =\frac{x^{12}-1}{x^2-1}\cdot (x^{12}+x)\\ & =(x^{10}+x^8+x^6+x^4+x^2+1)(x^{12}+x)\\ & =x^{22}+x^{20}+x^{18}+x^{16}+x^{14}+x^{12}+x^{11}+x^9+x^7+x^5+x^3+x.\end{array}$$This is a multiple of $q(x)$ that's very close to $p(x).$ In fact, when we take the difference, we get $$\begin{array}{rl}& p(x)-(x^{22}+x^{20}+x^{18}+x^{16}+x^{14}+x^{12}+x^{11}+x^9+x^7+x^5+x^3+x)\\ & =-x^{11}+x^{10}-x^9+x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+1.\end{array}$$Now, if we add $q(x),$ we get $$\begin{array}{rl}& p(x)+q(x)-(x^{22}+x^{20}+x^{18}+x^{16}+x^{14}+x^{12}+x^{11}+x^9+x^7+x^5+x^3+x)\\ & =2x^{10}+2x^8+2x^6+2x^4+2x^2+2.\end{array}$$We can also write this as $$\begin{array}{rl}& p(x)-(x^{22}+x^{20}+x^{18}+x^{16}+x^{14}+x^{12}+x^{11}+x^9+x^7+x^5+x^3+x-q(x))\\ & =2x^{10}+2x^8+2x^6+2x^4+2x^2+2.\end{array}$$So, we took $p(x),$ subtracted $$x^{22}+x^{20}+x^{18}+x^{16}+x^{14}+x^{12}+x^{11}+x^9+x^7+x^5+x^3+x-q(x),$$which we know is a multiple of $q(x),$ and ended up with $\boxed{2x^{10}+2x^8+2x^6+2x^4+2x^2+2}.$ Since the degree of this polynomial is less than the degree of $q(x),$ this is our remainder.