What is the remainder when $P(x^{12})$ is divided by $P(x)$? 
Let $P(x) = 1 + x + x^2 + x^3+ x^4 + x^5$. What is the remainder when $P(x^{12})$ is divided by $P(x)$?
  (a) $0$
  (b) $6$
  (c) $1+x$
  (d) $1 + x + x^2 + x^3+ x^4 $

The way I adopted to solve this question was by putting x=1 which gives remainder= $0$ and this matches with option (a). But the answer to this question is 6. I am not able to understand the solution given:

It would be great if someone could explain the solution in simpler or more intuitive terms. Another way of solving the question is also welcome.
 A: $P(x) = \frac{x^6-1}{x-1}$
$$x^{6n}-1 = (x^6-1)(1+x^6+x^{12}+\cdots +x^{6(n-1)})$$
Thus $$P(x^{12}) - 6 = x^{60} -1 + x^{48} - 1 + x^{36}-1 + x^{24}-1 + x^{12}- 1 $$ and hence $P(x^{12})-6$ is a multiple of $x^6-1$ and hence $P(x)$ divides $P(x^{12})-6$. Thus the reminder is 6.
A: mod $\,P(x)\,$ we have $\,\color{#c00}{x^6\equiv \bf 1}\,$ (not $\,x\equiv 1)\,$ by $\,\color{#c00}{x^6\!-1} = (x\!-\!1)P(x)\equiv 0$ 
so $\ P(x^{12}) = P((\color{#c00}{x^6})^2)\equiv P(\color{#c00}{\bf 1}^2)\equiv P(1)\equiv 6\ $ by the Polynomial Congruence Rule. 
A: Maybe this variant will be more intuitive:
Polynomial division leads to an equality 
$$P(x^{12})=P(x)Q(x)+R(x)$$
where the remainder $R(x)$ has degree $\le 4$ or $R(x)=0$
The roots of the divisor $P(x)$ are the non-real $5$-th roots of unity, and for any such root $\alpha$, we have
$$P(\alpha^{12})=P(\alpha)Q(\alpha)+R(\alpha)=R(\alpha).$$
One calculates that each $R(\alpha)$ is equal to $6$.
Now a theorem on polynomials asserts that a polynomial of degree $d$ is uniquely determined by its values at $d+1$ distinct points. So, $R(x)$, which coincides with the constant polynomial $6$ at $5$ points and has degree at most$4$, is necessarily equal to this constant.
