1. Lucie wrote two non-zero numbers on the board. Then she sequentially inserted the plus, minus, times, and division signs between them and correctly calculated all four problems. Among the results, there were only two different values. What two numbers could Lucie have written on the board?
SOLUTION. Let the sought numbers be $a, b$. Since $b \neq 0$, it is necessary that $a+b \neq a-b$. Each of the numbers $a \cdot b, a: b$ is therefore equal to either $a+b$ or $a-b$. It is therefore sufficient to analyze four cases and solve the system of equations in each of them. We will show a slightly more ingenious approach.

If it were true that

$$
a+b=a \cdot b \quad \text { and } \quad a-b=a: b \quad \text { or } \quad a+b=a: b \quad \text { and } \quad a-b=a \cdot b
$$

by multiplying the equalities, we would get $a^{2}-b^{2}=a^{2}$ in both cases, which contradicts the condition $b \neq 0$. Therefore, the numbers $a \cdot b$ and $a: b$ are either both equal to $a+b$, or both equal to $a-b$. In either case, it must be true that $a \cdot b=a: b$, from which, after rearrangement, $a\left(b^{2}-1\right)=0$. Since $a \neq 0$, necessarily $b \in\{1,-1\}$.

If $b=1$, the four results are $a+1, a-1, a, a$, which are three different values for any $a$.

For $b=-1$ we get the results $a-1, a+1, -a, -a$. These will be two different numbers precisely when $a-1=-a$ or $a+1=-a$. In the first case, we get $a=\frac{1}{2}$, in the second $a=-\frac{1}{2}$.

Lucie could have written on the board at the beginning either the numbers $\frac{1}{2}$ and -1, or the numbers $-\frac{1}{2}$ and -1.

## GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. We have three numbers with a sum of 2010, each of which is the arithmetic mean of the remaining two. What are these numbers? [We set up and solve a system of equations, the numbers must be the same, and they are therefore equal to 670.]

N2. We have three numbers, of which we know that each of them is the arithmetic mean of some two of our three numbers. Prove that our three numbers are the same. [Assume that one of our numbers is the average of itself and another of our numbers. Then we can denote them as $a, b$, $c$ such that $a=\frac{1}{2}(a+b)$. From this equation, it follows that $a=b$. The number $c$ is either the average of the numbers $a$ and $b$, from which we immediately have that it is equal to these numbers, or it is the average of itself and one of the numbers $a, b$, i.e., $c=\frac{1}{2}(c+a)$, from which we again get $c=a=b$. If each of our numbers is the arithmetic mean of the remaining two, we proceed similarly as in the previous problem.]

D1. Let $n$ be a natural number greater than 2. We have $n$ numbers with a sum of $n$, each of which is the arithmetic mean of the other numbers. What are these numbers? [Arrange our numbers in order, let $x_{1} \leqq x_{2} \leqq \ldots \leqq x_{n}$. The arithmetic mean of any group of numbers is not less than the smallest of them. The arithmetic mean of the numbers $x_{2}, x_{3}, \ldots, x_{n}$ is therefore at least $x_{2}$ and is equal to $x_{1}$ only in the case where none of the numbers $x_{3}, \ldots, x_{n}$ is greater than $x_{1}$. From this, we immediately get that all our numbers must be the same, and therefore equal to 1.]