The set $\{1,2, \ldots, 100\}$ has a subset $H$ with the property that the tenfold of any element in $H$ is not in $H$. What is the maximum number of elements $H$ can have?
Let's solve the task in a more general way. Let $G$ be a finite subset of positive integers, and $k$ a positive integer. The question can then be formulated as follows: what is the maximum number of elements that a subset $H$ of $G$ can have, such that no element of $H$ is a $k$-multiple of another element of $H$.

Let's call the elements of $G$ that do not have a $k$-multiple in $G$ the base numbers. From these, we can obtain all other elements by multiplying by an appropriate power of $k$. Thus, we can form chains in $G$ where the first position is a base number, followed by its $k-, k^{2}-, \ldots, k^{i}$-multiples. It is obvious that no two consecutive elements of such a chain can be in $H$ at the same time; in fact, this is equivalent to the desired condition for $H$. $H$ will then consist of the most elements if it contains as many numbers as possible from each chain.

For a chain of length $2l+1$, one of the 2nd and 3rd, 4th and 5th, ..., $2l$th and $(2l+1)$th numbers must definitely be omitted from $H$, so at most $l+1$ numbers can remain from the chain. However, so many can also remain: the 1st, 3rd, ..., $(2l+1)$th. For a chain of length $2l$, one of the 1st and 2nd, 3rd and 4th, ..., $(2l-1)$th and $2l$th must be omitted, so at most $l$ can remain, which is also achievable, for example, by selecting the 2nd, 4th, ..., $2l$th.

Let's calculate how many elements the thus constructed $H$ has. Let the lengths of the even-length chains be denoted by $l_{1}, l_{2}, \ldots, l_{s}$, and those of the odd-length chains by $t_{1}, t_{2}, \ldots, t_{r}$. Then the number of elements of $H$ is

$$
|H|=\frac{l_{1}}{2}+\frac{l_{2}}{2}+\cdots+\frac{l_{s}}{2}+\frac{t_{1}+1}{2}+\frac{t_{2}+1}{2}+\cdots+\frac{t_{r}+1}{2}
$$

However, the number of elements of $G$ is

$$
|G|=l_{1}+l_{2}+\cdots+l_{s}+t_{1}+t_{2}+\cdots+t_{r}
$$

so the number of elements of $H$ is

$$
|H|=\frac{l_{1}+l_{2}+\cdots+l_{s}+t_{1}+t_{2}+\cdots+t_{r}}{2}+\frac{r}{2}=\frac{|G|+r}{2}
$$

Now let's consider our original problem, where $k=10, G=\{1,2, \ldots, 100\}$. Let's determine the chains:

1) the $1-10-100$ is a three-element chain
2) the $2-20 ; 3-30 ; \ldots 9-90$ are two-element chains, a total of 8
3) the remaining $100-16-3=81$ numbers form one-element chains.

So $|G|=100, r=81+1=82$, hence the answer to the question is $\frac{182}{2}=91$. An example of an $H$ set is $\{1,2, \ldots, 100\}-\{2,3, \ldots, 10\}$

Based on the work of Péter Burcsi (Pápa, Türr István Gymnasium, 2nd grade)

Note. In the reasoning for the general case, we provided an $H$ for which we saw that there is no $H'$ with more elements that also satisfies the conditions. However, it is not said that an $H'$ of the same size as it does not exist. The selection from odd-length chains is unique, but there are multiple possibilities for even-length chains. For example, instead of the 2nd, 4th, ..., $2l$th, the 1st, 3rd, ..., $(2l-1)$th; or the 1st, 4th, 6th, ..., $2l$th. Generally, from a $2l$-length chain, this can be done in $(l+1)$ ways. This can be proven by induction. For $l=1$, it is obvious: either the 1st or the 2nd can be taken. Let's examine from $n$ to $n+1$. If we chose the $(2n+1)$th, only the $(2n-1)$th, $(2n-3)$th, ..., 3rd, 1st can precede it, which is 1 possibility. If we chose the $(2n+2)$th, then from the $1$st, $2$nd, ..., $2n$th, the elements can be taken in $(n+1)$ ways according to the case $l=n$. Since we had to choose one from the $(2n+1$th, $2n+2$th) pair, this is all the possibilities, i.e., $n+2$, exactly as we stated.