smc

Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$. Then $r(x)$ is the unique polynomial such that $$g(x^{12})-r(x)=x^{60}+x^{48}+x^{36}+x^{24}+x^{12}+1-r(x)$$ is divisible by $g(x)=x^5+x^4+x^3+x^2+x+1$, and $\mathrm{deg}r(x)<5$. Note that $(x-1)(x^5+x^4+x^3+x^2+1)=x^6-1$ is a multiple of $g(x)$. Also, $$\begin{gathered} g(x^{12})-6=x^{60}+x^{48}+x^{36}+x^{24}+x^{12}-5 \\ =(x^{60}-1)+(x^{48}-1)+(x^{36}-1)+(x^{24}-1)+(x^{12}-1). \end{gathered}$$ Each term is a multiple of $x^6-1$. For example, $$x^{60}-1=(x^6-1)(x^{54}+x^{48}+\cdots +x^6+1).$$ Hence, $g(x^{12})-6$ is a multiple of $x^6-1$, which means that $g(x^{12})-6$ is a multiple of $g(x)$. Therefore, the remainder is $\boxed{6}$. The answer is (A).