2aras Let $X=\{1,2, \cdots, 100\}, A$ be a subset of $X$, such that for any two elements $x, y \in A (x<y)$, we have $y \neq 3x$. Find the maximum value of $|A|$.


Let $A=\{3 k+1 \mid k=0,1,2, \cdots, 33\} \cup\{3 k+2 \mid k=0,1,2, \cdots, 32\}$ $\cup\{9,18,36,45,63,72,81,90,99\}$, then $A$ clearly meets the conditions, and at this point $|A|=76$.

On the other hand, consider 24 sets $A_{k}=\{k, 3 k\}(k=1,2,12,13, \cdots, 33)$, which contain 48 distinct numbers. Excluding these numbers, there are still 52 numbers in $X$. Each of these 52 numbers can be made into a single-element set, together with the previous 24 sets, making a total of $52+24=76$ sets. If $|A|>76$, then $A$ must contain 2 numbers from some set $A_{k}$, where the larger number is three times the smaller number, which is a contradiction.
In summary, the maximum value of $|A|$ is 76.
Comment We first construct a set $A$ that meets the conditions. To ensure that $y \neq 3 x$, a sufficient condition is that $A$ contains no multiples of 3. Therefore, all numbers that are not multiples of 3 can belong to $A$. Further, we find that $A$ can also contain some "multiples of 3," provided these "multiples of 3" meet two conditions: one is that they are not three times the aforementioned numbers, which only requires that they remain multiples of 3 after being divided by 3, i.e., they are multiples of 9; the other is that there is no threefold relationship among these numbers. Thus, we can select $9, 18, 36, 45, 63, 72, 81, 90, 99$ to belong to $A$, at which point $|A|=76$. Therefore, we can conjecture that the maximum value of $|A|$ is 76.