G10.2Let $f(x)=x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1$. When $f\left(x^{10}\right)$ is divided by $f(x)$, the remainder is $b$. Find the value of $b$.
Consider the roots of $f(x)=x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1=0$
$f(x)$ can be rewritten as $\frac{x^{10}-1}{x-1}=0$, where $x \neq 1$
There are 9 roots $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}, \alpha_{5}, \alpha_{6}, \alpha_{7}, \alpha_{8}, \alpha_{9}$ : where $\alpha_{1}{ }^{10}=1$ and $\alpha_{i} \neq 1$ for $1 \leq i \leq 9$
Let $f\left(x^{10}\right)=f(x) Q(x)+a_{8} x^{8}+a_{7} x^{7}+a_{6} x^{6}+a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$.
$f\left(\alpha_{\mathrm{i}}^{10}\right)=f\left(\alpha_{\mathrm{i}}\right) Q\left(\alpha_{\mathrm{i}}\right)+a_{8} \alpha_{\mathrm{i}}^{8}+a_{7} \alpha_{\mathrm{i}}^{7}+a_{6} \alpha_{\mathrm{i}}^{6}+a_{5} \alpha_{\mathrm{i}}^{5}+a_{4} \alpha_{\mathrm{i}}^{4}+a_{3} \alpha_{\mathrm{i}}^{3}+a_{2} \alpha_{\mathrm{i}}^{2}+a_{1} \alpha_{\mathrm{i}}+a_{0}, 1 \leq i \leq 9$
$f(1)=0 \cdot Q\left(\alpha_{\mathrm{i}}\right)+a_{8} \alpha_{\mathrm{i}}^{8}+a_{7} \alpha_{\mathrm{i}}^{7}+a_{6} \alpha_{\mathrm{i}}^{6}+a_{5} \alpha_{\mathrm{i}}^{5}+a_{4} \alpha_{\mathrm{i}}^{4}+a_{3} \alpha_{\mathrm{i}}^{3}+a_{2} \alpha_{\mathrm{i}}^{2}+a_{1} \alpha_{\mathrm{i}}+a_{0}$ for $1 \leq i \leq 9$
$a_{8} \alpha_{\mathrm{i}}^{8}+a_{7} \alpha_{\mathrm{i}}^{7}+a_{6} \alpha_{\mathrm{i}}^{6}+a_{5} \alpha_{\mathrm{i}}^{5}+a_{4} \alpha_{\mathrm{i}}^{4}+a_{3} \alpha_{\mathrm{i}}^{3}+a_{2} \alpha_{\mathrm{i}}^{2}+a_{1} \alpha_{\mathrm{i}}+a_{0}=10$ for $1 \leq i \leq 9$
$\therefore \alpha_{i}(1 \leq i \leq 9)$ are the roots of $a_{8} x^{8}+a_{7} x^{7}+a_{6} x^{6}+a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}-10=0$
Since a polynomial of degree 8 has at most 8 roots and it is satisfied by $\alpha_{\mathrm{i}}$ for $1 \leq i \leq 9$.
$\therefore a_{8} x^{8}+a_{7} x^{7}+a_{6} x^{6}+a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}-10$ must be a zero polynomial.
$a_{8}=0, a_{7}=0, a_{6}=0, a_{5}=0, a_{4}=0, a_{3}=0, a_{2}=0, a_{1}=0, a_{0}=10$
The remainder when $f\left(x^{10}\right)$ is divided by $f(x)$ is $a_{0}=10$.
Method 2 (Provided by Pui Ching Middle School 李國柱老師)
$$
\begin{aligned}
f\left(x^{10}\right) & =x^{90}+x^{80}+x^{70}+x^{60}+x^{50}+x^{40}+x^{30}+x^{20}+x^{10}+1 \\
& =x^{90}-1+x^{80}-1+x^{70}-1+x^{60}-1+x^{50}-1+x^{40}-1+x^{30}-1+x^{20}-1+x^{10}-1+10 \\
& =\left(x^{10}-1\right) g_{1}(x)+\left(x^{10}-1\right) g_{2}(x)+\left(x^{10}-1\right) g_{3}(x)+\ldots+\left(x^{10}-1\right) g_{9}(x)+10
\end{aligned}
$$
where $g_{1}(x)=x^{80}+x^{70}+\ldots+x^{10}+1, g_{2}(x)=x^{70}+x^{60}+\ldots+x^{10}+1, \ldots, g_{9}(x)=1$
$$
\begin{aligned}
f\left(x^{10}\right) & =\left(x^{10}-1\right)\left[g_{1}(x)+g_{2}(x)+\ldots+g_{9}(x)\right]+10 \\
& =f(x)(x-1)\left[g_{1}(x)+g_{2}(x)+\ldots+g_{9}(x)\right]+10
\end{aligned}
$$

The remainder is 10 .
Method 3 (Provided by Pui Ching Middle School 李國柱老師)
Clearly $f(1)=10$.
By division algorithm, $f(x)=(x-1) Q(x)+10$, where $Q(x)$ is a polynomial
$$
\begin{aligned}
f\left(x^{10}\right) & =\left(x^{10}-1\right) Q\left(x^{10}\right)+10 \\
& =(x-1)\left(x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\right) Q\left(x^{10}\right)+10 \\
& =f(x)(x-1) Q\left(x^{10}\right)+10
\end{aligned}
$$

The remainder is 10 .