4. Determine the smallest natural number $n$ for which there exists a set

$$
M \subset\{1,2, \ldots, 100\}
$$

of $n$ elements that satisfies the conditions:
a) 1 and 100 belong to the set $M$,

b) for every $a \in M \backslash\{1\}$, there exist $x, y \in M$ such that $a=x+y$.
Solution. Let $M=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\} \subset\{1,2, \ldots, 100\}$, where

$$
1=a_{1}8$, or $n \geq 9$. If $a_{k}=50$ for some index $k$, then from $a_{5}+a_{6} \leq 487$, i.e., $k \geq 8$, so $n \geq 9$. It is straightforward to verify that for the sets

$$
\begin{aligned}
& M_{1}=\{1,2,4,6,10,20,40,60,100\} \\
& M_{2}=\{1,2,3,6,12,13,25,50,100\}
\end{aligned}
$$

all conditions of the problem are satisfied. Therefore, the required number is equal to 9.

## III and IV year