How many integers $a$ divide $2^{4} \times 3^{2} \times 5$?
It should be noted here that a divisor $n$ of $2^{4} \times 3^{2} \times 5$ is of the form $n=2^{x} \times 3^{y} \times 5^{z}$. Indeed, if a prime number other than 2, 3, or 5 divided $n$, then it would also divide $2^{4} \times 3^{2} \times 5$, which is not possible. It is also necessary that $x \leqslant 4, y \leqslant 2$, and $z \leqslant 1$, and this will give us all possible divisors. In summary, to create a divisor, we need to

1. Choose $x$ between 0 and 4, so 5 possibilities,
2. Choose $y$ between 0 and 2, so 3 possibilities,
3. Choose $z$ between 0 and 1, so 2 possibilities,

This gives a total of $5 \times 3 \times 2=30$ possible divisors.

## Proposition 2.

For a set of size $n$, there are $2^{n}$ "subsets".

Proof. The reasoning is as follows: consider the first element, you have the choice to keep it or leave it out, that is, 2 possibilities. The same for the second element, either you keep it or you leave it out. etc... In total, this gives you $2 \times 2 \times \cdots \times 2=2^{n}$ possibilities.