New Zealand Mathematical Olympiad

Solution:
For each $n = 1, 2, \ldots, 9$ there are $\lfloor 9 / n \rfloor$ different possibilities for the $n^{\mathrm{th}}$ digit. For example, there are $\lfloor 9 / 2 \rfloor = 4$ possible choices for the second digit (these being $2, 4, 6$ and $8$). Therefore the answer is
$$\left\lfloor \frac{9}{1} \right\rfloor \times \left\lfloor \frac{9}{2} \right\rfloor \times \left\lfloor \frac{9}{3} \right\rfloor \times \left\lfloor \frac{9}{4} \right\rfloor \times \left\lfloor \frac{9}{5} \right\rfloor \times \left\lfloor \frac{9}{6} \right\rfloor \times \left\lfloor \frac{9}{7} \right\rfloor \times \left\lfloor \frac{9}{8} \right\rfloor \times \left\lfloor \frac{9}{9} \right\rfloor$$
which equals $9 \times 4 \times 3 \times 2 \times 1 \times 1 \times 1 \times 1 \times 1 = 216$