Problem 4. Let $n$ be a non-zero natural number. We will say that a set $A$ of natural numbers is complete of size $n$ if its elements are non-zero, and the set of all remainders obtained by dividing an element of $A$ by an element of $A$ is $\{0,1,2, \ldots, n\}$. For example, the set $\{3,4,5\}$ is a complete set of size 4.

Determine the minimum number of elements of a complete set of size 100.
Solution. Answer: 27.

An example of a complete set of size 100 with 27 elements is

$$
\{76,77,78, \ldots, 100\} \cup\{51,152\}
$$

Indeed, for the divisions $100: x, 76 \leq x \leq 100$, we obtain the remainders $0,1,2, \ldots, 24$, for the divisions $x: 51, 76 \leq x \leq 100$, we obtain the remainders $25,26, \ldots, 49$, for the divisions $152: x, 77 \leq x \leq 100$, we obtain the remainders $52,53, \ldots, 75$, for the divisions $x: 152, 76 \leq x \leq 100$, we obtain the remainders $76,77, \ldots, 100$, for the division 51 : 152 we obtain the remainder 51 and for the division 152 : 51 we obtain the remainder 50. ............... 4p

We now show that any complete set of size 100 has at least 27 elements.

We observe that if $A=\left\{a_{1}100$, it follows that $A$ has at least 27 elements. 3 p

## National Mathematics Olympiad  National Stage, Satu Mare, April 4, 2018  SOLUTIONS AND ORIENTATIVE SCORING - GRADE 7