NZMO Round One

Solution:

For any positive integer $n$, the number of multiples of $n$ less than or equal to $2019$ is given by
$$\left\lfloor \frac{2019}{n} \right\rfloor.$$
So there are $\left\lfloor \frac{2019}{18} \right\rfloor = 112$ multiples of $18$, and $\left\lfloor \frac{2019}{21} \right\rfloor = 96$ multiples of $21$. Moreover, since $\operatorname{lcm}(18,21) = 126$ there are $\left\lfloor \frac{2019}{126} \right\rfloor = 16$ positive integers less than $2019$ which are a multiple of both $18$ and $21$.

Therefore the final answer is
$$\left\lfloor \frac{2019}{18} \right\rfloor + \left\lfloor \frac{2019}{21} \right\rfloor - 2\left\lfloor \frac{2019}{126} \right\rfloor = 112 + 96 - 2 \times 16 = 176.$$