2. Determine how many numbers we can select $z$ from the set $\{1,2,3, \ldots, 75599,75600\}$ such that the set includes the number 75600 and for any two selected numbers $a$, $b$, it holds that $a$ is a divisor of $b$ or $b$ is a divisor of $a$. (List all possibilities.)
SOLUTION. Let's consider a set M that satisfies the conditions given in the problem. Since M contains the number 75600, it must be at least a one-element set. Further, notice that if we remove some number $a \neq 75600$ from the set M, we get a set $\mathrm{M}^{\prime} \subset \mathrm{M}$, which also satisfies the given conditions. Let's verify this. The set $\mathrm{M}^{\prime}$ still contains the number 75600. If $x, y$ are any two numbers from the set $\mathrm{M}^{\prime}$, it automatically holds that $x \mid y$ or $y \mid x$, because this holds for them as elements of the set $\mathrm{M}$.

By this, we have essentially proven that if we find a set with $m$ elements that satisfies the conditions of the problem, then there exists a $k$-element set with the required properties for any $k, 1 \leqq k \leqq m$. Therefore, it is sufficient to find a suitable set that has the maximum possible number of elements.

If $a$ is any element of the set M, it is primarily $a \leqq 75600$. If $a<75600$, according to the problem, it must hold that $a \mid 75600$. The set M therefore contains only divisors of the number 75600.

The prime factorization of the number 75600 is $75600=2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7$. Every divisor of the number 75600 thus has the form $2^{\alpha} \cdot 3^{\beta} \cdot 5^{\gamma} \cdot 7^{\delta}$, where $0 \leqq \alpha \leqq 4,0 \leqq \beta \leqq 3,0 \leqq \gamma \leqq 2,0 \leqq \delta \leqq 1$. Each element of $\mathrm{M}$ is therefore characterized by an ordered quadruple $(\alpha, \beta, \gamma, \delta)$ of the corresponding exponents in the given prime factorization. If $p$ and $p^{\prime}$ are two different elements of $\mathrm{M}$ and, for example, $p<p^{\prime}$, then according to the problem, it must simultaneously hold that $\alpha \leqq \alpha^{\prime}, \beta \leqq \beta^{\prime}, \gamma \leqq \gamma^{\prime}, \delta \leqq \delta^{\prime}$, and at least one inequality must be strict (otherwise it would hold that $p=p^{\prime}$), from which the inequality $\alpha+\beta+\gamma+\delta<\alpha^{\prime}+\beta^{\prime}+\gamma^{\prime}+\delta^{\prime}$ follows. Since in our case $0 \leqq \alpha+\beta+\gamma+\delta \leqq 10$, the set $M$ can contain at most 11 elements. Such a set is, for example,

$$
\mathrm{D}=\left\{1,2,2^{2}, 2^{3}, 2^{4}, 2^{4} \cdot 3,2^{4} \cdot 3^{2}, 2^{4} \cdot 3^{3}, 2^{4} \cdot 3^{3} \cdot 5,2^{4} \cdot 3^{3} \cdot 5^{2}, 2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7\right\}
$$

By this, we have proven that from the given set, we can select the required number of elements $1,2, \ldots, 11$ (including the number 75600).