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.

$P(x) = \frac{x^6-1}{x-1}$
$$x^{6n}-1 = (x^6-1)(1+x^6+x^{12}+\cdots +x^{6(n-1)})$$
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.
