Finale

Notice that: $x^{6}-1=(x-1)g(x).$ Therefore $g(x)$ is a factor of $x^{6}-1$ , so we can write: $x^{6} \equiv 1 \ (mod \ g(x)).$ Hence: $g(x^{12})=\sum_{n=0}^{5}{(x^{12})^{n}} = \sum_{n=0}^{5}{(x^{6})^{2n}} \equiv \sum_{n=0}^{5}{(1)^{2n}} \ (mod \ g(x))$ $\Rightarrow g(x^{12}) \equiv \boxed{6} \ (mod \ g(x)).$

$g(x) = x^{5}+x^{4}+x^{3}+x^{2}+x+1$

$g(x^{12}) = x^{60}+x^{48}+x^{36}+x^{24}+x^{12}+1$

$g(x^{12}) = (x^{60}-1)+(x^{48}-1)+(x^{36}-1)+(x^{24}-1)+(x^{12}-1)+6$

Notice that $(x^{60}-1)$ is divisible by $(x^{6}-1)$ .

This is also true for $(x^{48}-1)$ , $(x^{36}-1)$ , $(x^{24}-1)$ and $(x^{12}-1)$ .

Also notice that $(x^{6}-1)$ is divisible by $g(x)$ .

Hence, the remainder when $g(x^{12})$ is divided by $g(x)$ is $6$ .

All solutions are great but this is......

$let\quad the\quad remainder\quad is\quad B\\ \\ \cfrac { g\left( { x }^{ 12 } \right) }{ g\left( x \right) } =A+\cfrac { B }{ g\left( x \right) } \\ \\ \Rightarrow g\left( { x }^{ 12 } \right) =Ag\left( x \right) +B\\ \\ as\quad g\left( -1 \right) =0\\ \\ so,B={ lim }_{ x\rightarrow -1 }\left( Ag\left( x \right) +B \right) \\ \\ \quad ={ lim }_{ x\rightarrow -1 }g\left( { x }^{ 12 } \right) \\ \\ \quad =g\left( 1 \right) =6$

Let $ω=e^{\frac{2πi}{6}}$ be the sixth root of unity.

The polynomial $g(x)=1+x+x^2+x^3+x^4+x^5$ has roots $ω^{k}$ where $k=1,2,..,5$

Let $g(x^{12})=p(x)g(x)+t(x)$ where $degt(x)≤4$ .

Now, $6=g(1)=g(ω^{6})=g((ω^{6})^{2r})=g((ω^{r})^{12})=p(ω^{r})g(ω^{r})+t(ω^{r})$ for $r=1,..,5$ .

Since, $g(ω^{r})=0$ for $r=1,2,..,5$ , $t(ω^{r})-6=0$

Hence, $t(x)-6=0$ for $5$ values of $x$ .But degree of $t(x)$ is at most $4$

So, $t(x)=6$ for all values of $x$ .Hence, the remainder is $6$ .