Find the remainder of the division of the polynomial $P(x)=x^{6 n}+x^{5 n}+x^{4 n}+x^{3 n}+x^{2 n}+x^{n}+1$ by $Q(x)=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}$ $+x+1$, given that $n$ is a multiple of 7.
The difference $x^{k n}-1$ is divisible by $x^{7}-1$, and therefore also by $Q(x)$. Therefore, $P(x)-7=\left(x^{6 n}-1\right)+\left(x^{5 n}-1\right)+\ldots+\left(x^{n}-1\right)$ is divisible by $Q(x)$.

## Answer

7.