4. Let the faces of the cube be denoted by $S_{1}, S_{2}, \ldots, S_{6}$ such that face $S_{1}$ is opposite to face $S_{6}$, face $S_{2}$ is opposite to $S_{5}$, and $S_{3}$ is opposite to $S_{4}$. Let the number on face $S_{i}$ be denoted by $c_{i}$. Clearly, any vertex of the cube belongs to exactly one pair of opposite faces. This means that in each step, the value of the sums $c_{1}+c_{6}, c_{2}+c_{5}$, and $c_{3}+c_{4}$ of the numbers on opposite faces increases by 1. Therefore, if at the end it is to hold that $c_{1}=c_{2}=c_{3}=c_{4}=c_{5}=c_{6}$, and thus also

$$
c_{1}+c_{6}=c_{2}+c_{5}=c_{3}+c_{4},
$$

the sums of the numbers on opposite faces of the cube must be the same at the beginning (and will remain the same after each step).

We will show that condition (1) is also sufficient. Let the numbers on the faces of the cube satisfy (1). We will describe a sequence of steps after which the numbers on all faces of the cube will be the same. Let a step in which we increase the numbers on faces $S_{i}, S_{j}, S_{m}$ be denoted by $k_{i j m}$. Without loss of generality, let $c_{1}=p$ be the largest of the six numbers on the cube. Now we perform $\left(p-c_{2}\right)$ times the step $k_{246}$ and $\left(p-c_{3}\right)$ times the step $k_{356}$. This will ensure that the numbers on faces $S_{1}, S_{2}, S_{3}$ are the same, namely $p$. Due to condition (1), the numbers on faces $S_{4}, S_{5}, S_{6}$ will also be the same, let their value be $q$. If $p \neq q$, it is now sufficient to perform $(p-q)$ times the step $k_{456}$ if $p>q$, or $(q-p)$ times the step $k_{123}$ if $q>p$.

Our task is to determine the number of such sets $\mathrm{M}=\left\{c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}\right\}$ of distinct natural numbers for which

$$
c_{1}+c_{2}+c_{3}+c_{4}+c_{5}+c_{6}=60 \quad \text { and } \quad c_{1}+c_{6}=c_{2}+c_{5}=c_{3}+c_{4} .
$$

From this, it follows that $3\left(c_{1}+c_{6}\right)=60$, thus

$$
c_{1}+c_{6}=c_{2}+c_{5}=c_{3}+c_{4}=20 .
$$

Without loss of generality, we can assume that $c_{1}<c_{2}<c_{3}<c_{4}<c_{5}<c_{6}$, or (considering the equalities (2))

$$
c_{1}<c_{2}<c_{3}<10<c_{4}<c_{5}<c_{6} .
$$

For each triplet $\left(c_{1}, c_{2}, c_{3}\right)$ satisfying $c_{1}<c_{2}<c_{3}<10$, the remaining numbers $c_{4}, c_{5}, c_{6}$ can be calculated from (2). The number of all suitable sets $\mathrm{M}$ is therefore equal to the number of different triplets of natural numbers $\left(c_{1}, c_{2}, c_{3}\right)$ that satisfy the condition $c_{1}<c_{2}<c_{3}<10$, which is

$$
\left(\begin{array}{l}
9 \\
3
\end{array}\right)=\frac{9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 3}=84
$$

For a complete solution, award 6 points, of which 1 point for finding and correctly justifying the necessary condition (1), 3 points for proving that (1) is also a sufficient condition, and 2 points for determining the number of sets M.
4. Let M be a set of six distinct positive integers whose sum is 60. We will write all of them on the faces of a cube, with exactly one on each face. In one step, we choose any three faces of the cube that share a common vertex and increase each of the numbers on these three faces by 1. Determine the number of all such sets M whose numbers can be written on the faces of the cube in such a way that after a finite number of appropriate steps, the numbers on all faces will be the same.

The regional round of category A takes place

## on Tuesday, January 18, 2011

so that it starts in the morning and the contestants have 4 hours of clean time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. Allowed aids are writing and drawing tools, school MF tables, and calculators without a graphical display. These details will be communicated to the students before the start of the competition.