1. A non-empty set of natural numbers is called small if it has fewer elements than its smallest element. Determine the number of all small sets $\mathrm{M}$ that are subsets of the set $\{1,2,3, \ldots, 100\}$ and have the following property: if two different numbers $x$ and $y$ belong to $\mathrm{M}$, then the number $|x-y|$ also belongs to $\mathrm{M}$.
SOLUTION. First, we will determine what all non-empty finite sets $M$ of natural numbers with the (key) property from the conclusion of the problem look like. Only then will we assess which of these sets are small, and determine the number of those that are composed of numbers from 1 to 100.

Let $\mathrm{M}$ be any non-empty finite set of natural numbers with the property: if $x, y \in M$ and $x \neq y$, then $|x-y| \in \mathrm{M}$. Assume that M has exactly $k$ elements, and order them from the smallest number to the largest:

$$
x_{1}<x_{2}<\ldots<x_{k}
$$

If $k=1$, then $M=\{x_{1}\}$, and the property is trivially satisfied. If $k>1$, then the number $x_{2}-x_{1}=\left|x_{2}-x_{1}\right|$ must belong to $\mathrm{M}$ according to the property and be smaller than $x_{2}$, so it must equal $x_{1}$. From the equality $x_{2}-x_{1}=x_{1}$, we get $x_{2}=2 x_{1}$. Similarly, the numbers $x_{3}-x_{2}, x_{3}-x_{1}$ are two numbers from $\mathrm{M}$ that are smaller than $x_{3}$, and $x_{3}-x_{2}<x_{3}-x_{1}$. Therefore, $x_{3}-x_{2}=x_{1}$ and $x_{3}-x_{1}=2 x_{1}$, which gives $x_{3}=3 x_{1}$. By induction, we can show that $x_{i}=i x_{1}$ for all $i=1,2, \ldots, k$.

Thus, the set $M$ must be of the form $\{x, 2x, 3x, \ldots, kx\}$ for some natural number $x$. The set $M$ is small if and only if $kx \leq 100$. For a given $x$, the number of such sets is the largest integer $k$ such that $kx \leq 100$, which is $\left\lfloor \frac{100}{x} \right\rfloor$. Summing over all possible values of $x$ from 1 to 100, the total number of small sets is:

$$
\sum_{x=1}^{100} \left\lfloor \frac{100}{x} \right\rfloor
$$

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