19. How many pairs of numbers $(m, n)$ are there such that the following statement is true?
'A regular $m$-sided polygon has an exterior angle of size $n^{\circ}$ and a regular $n$-sided polygon has an exterior angle of size $m^{\circ}$.
A 24
B 22
C 20
D 18
E 16
Solution
C
We first note that a polygon has at least 3 sides, so we need consider only cases where $m \geq 3$ and $n \geq 3$.
The exterior angle of a regular $m$-sided polygon is $\left(\frac{360}{m}\right)^{\circ}$. Hence, the condition for a regular $m$-sided polygon to have an exterior angle of $n^{\circ}$ is that $n=\frac{360}{m}$. This condition may be rewritten as $m n=360$.
Similarly, this is the condition that a regular $n$-sided polygon has an exterior angle of $m^{\circ}$.
Therefore the pair of numbers ( $m, n$ ) satisfies the condition of the question if, and only if $m$ and $n$ are both at least 3 and $m n=360$.

There are 10 ways of expressing 360 as the product of positive integers which are both at least 3 . These are
$$
3 \times 120,4 \times 90,5 \times 72,6 \times 60,8 \times 45,9 \times 40,10 \times 36,12 \times 30,15 \times 24 \text { and } 18 \times 20 .
$$

Each of these 10 factorizations can be taken in either order to give two pairs $(m, n)$ which meet the required condition. For example, corresponding to the factorization $3 \times 120$ we see that $(m, n)$ may be either $(3,120)$ or $(120,3)$.
It follows that there are $10 \times 2$, that is, 20 pairs of numbers $(m, n)$ meeting the required condition.