如图所示，若干个正 n 边形首尾相接排成一个环，围成一个正 N 边形。

1. Write the interior angles of the small and large regular polygons.

Interior angle of each regular $n$-sided polygon:
$\frac{(n-2)180^\circ}{n}$
Interior angle of the regular $N$-sided polygon:
$\frac{(N-2)180^\circ}{N}$

2. Form an angle equation at a vertex of the large polygon.

At each vertex, two interior angles of the smaller polygons and one interior angle of the large polygon add to $360^\circ$.

$2\left(\frac{(n-2)180}{n}\right)+\frac{(N-2)180}{N}=360$

3. Simplify the equation to express $N$ in terms of $n$.

Divide throughout by $180$:
$\frac{2(n-2)}{n}+\frac{N-2}{N}=2$

Since
$\frac{N-2}{N}=1-\frac{2}{N},$
we get
$\frac{2n-4}{n}+1-\frac{2}{N}=2$

$2-\frac{4}{n}+1-\frac{2}{N}=2$

$1-\frac{4}{n}=\frac{2}{N}$

$\frac{n-4}{n}=\frac{2}{N}$

Cross-multiplying,
$N(n-4)=2n$

$N=\frac{2n}{n-4}$

1. Substitute $N=8$ into the formula from part (a).

From part (a),
$N=\frac{2n}{n-4}.$
For a regular octagon, $N=8$:
$8=\frac{2n}{n-4}.$

2. Solve for $n$ and interpret whether it is possible.

$8(n-4)=2n$
$8n-32=2n$
$6n=32$
$n=\frac{16}{3}=5\frac{1}{3}.$
This is not an integer, so it cannot be the number of sides of a regular polygon.