What is the maximum number of elements of a subset of $\{1,2, \ldots, 100\}$ such that none of its elements is a multiple of any other?
Initially, observe that the set $\{51,52,53, \ldots, 100\}$ has 50 elements and none of its elements is a multiple of another. Thus, the subset with the largest number of elements that satisfies the required property has at least 50 elements. To conclude that 50 is the largest possible number of elements of a subset that satisfies the required property, it suffices to show that every subset with more than 50 elements contains two multiples. For this, we denote $B=\{1,2,3, \ldots, 100\}$ and divide $B$ into 50 disjoint subsets, considering the various subsets of $B$ whose elements are of the form odd number $\times 2^{k}$, with $k$ a natural number. Since there are only 50 odd numbers between 1 and 100, we obtain fifty pairwise disjoint subsets constructed in this way, as follows.

- $A_{1}=\{1,2,4,8,16,32,64\}=B \cap\left\{1 \times 2^{k} \mid\right.$ for some $\left.k=0,1,2,3, \ldots\right\}$;
- $A_{2}=\{3,6,12,24,48,96\}=B \cap\left\{3 \times 2^{k} \mid\right.$ for some $\left.k=0,1,2,3, \ldots\right\}$;
- $A_{3}=\{5,10,20,40,80\}=B \cap\left\{5 \times 2^{k} \mid\right.$ for some $\left.k=0,1,2,3, \ldots\right\}$;
- $A_{50}=\{99\}=B \cap\left\{99 \times 2^{k} \mid\right.$ for some $\left.k=0,1,2,3, \ldots\right\}$.

Observe that $B$ is the union of these fifty subsets, that is,

$$
B=\{1,2, \ldots, 100\}=A_{1} \cup A_{2} \cup \ldots \cup A_{50}
$$

and that if two elements of $B$ are in the same subset $A_{i}$, then one of them is a multiple of the other. Thus, if a subset $A$ of $B$ has more than 50 elements, we can assert that there are at least two elements of $A$ in the same subset $A_{i}$ and, therefore, one of them is a multiple of the other. This proves that 50 is the maximum number of elements of any subset of $B$ that does not have two elements such that one is a multiple of the other.