1. A regular polygon $P$ has four more sides than another regular polygon $Q$, and their interior angles differ by $1^{\circ}$. How many sides does $P$ have?
Solution 1
Say $P$ has $n$ sides. Then $Q$ has $(n-4)$ sides.
Given that their interior angles differ by $1^{\circ}$, their exterior angles also differ by $1^{\circ}$.
Using $p$ to represent the exterior angle of $P$ and $q$ to represent the exterior angle of $Q$, we have
$$
p=\frac{360}{n} \quad \text { and } \quad q=\frac{360}{n-4} .
$$

Since $P$ has more sides than $Q, p<q$, so $p+1=q$. Hence
$$
\frac{360}{n}+1=\frac{360}{n-4} \text {. }
$$

Multiplying through by $n(n-4)$ gives
$$
360(n-4)+n(n-4)=360 n \text {, }
$$
which simplifies to
$$
n^{2}-4 n-1440=0 \text {. }
$$

This factorises to give
$$
(n+36)(n-40)=0,
$$
which (since $n$ cannot be negative) leads to $n=40$. Hence $P$ has 40 sides (and $Q$ has 36).
Solution 2
Defining $n$ as before, we can work directly with the interior angles of $P$ and $Q$ (noting that the interior angle of $P$ is larger than the interior angle of $\mathrm{Q}$ :
$$
\begin{aligned}
\frac{180(n-2)}{n} & =\frac{180(n-6)}{n-4}+1 \\
180(n-2)(n-4) & =180 n(n-6)+n(n-4) \\
180 n^{2}-1080 n+1440 & =180 n^{2}-1080 n+n^{2}-4 n,
\end{aligned}
$$
which leads to the same quadratic equation as in solution 1 .
REMARK
The correct factorisation of the quadratic is not immediately obvious. One route to spotting the factors required is to consider how we reached 1440 in the first place: $1440=360 \times 4=36 \times 40$, which leads to the factorisation straight away.