
4. Let $M$ be a set of six mutually different positive integers which sum up to 60 . We write these numbers on faces of a cube (on each face one). In a move we choose three faces with a common vertex and we increase each number on these faces by one. Find the number of all sets $M$, whose elements (numbers) can be written on the faces of the cube in such a way that we can even up the numbers on the faces in finitely many moves.

(Peter Novotný)


Solution. Let us denote the faces of the cube by $S_{1}, S_{2}, \ldots, S_{6}$, where $S_{1}$ is opposite to $S_{6}$, and $S_{2}$ opposite to $S_{5}$. Let $c_{i}$ be written on $S_{i}$. Since any vertex of the cube lies just in one of any pair of opposite faces, we increase by one the sums $c_{1}+c_{6}$, $c_{2}+c_{5}$ a $c_{3}+c_{4}$. Should at the end be $c_{1}=c_{2}=c_{3}=c_{4}=c_{5}=c_{6}$, and so

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

the sums of numbers on the opposite faces have to be the same already at the beginning (and after each move).

We show that (1) is also a sufficient condition. Let the numbers on faces of the cube satisfy (1). Let $k_{i j m}$ be the move, in which we increase the numbers on $S_{i}, S_{j}$, $S_{m}$. Wlog we may assume that $c_{1}=p$ is the maximal number on the cube. We make $\left(p-c_{2}\right)$ times move $k_{246}$ and $\left(p-c_{3}\right)$ times move $k_{356}$ after which the numbers on faces $S_{1}, S_{2}, S_{3}$ will be the same. Due to (1) the numbers on faces $S_{4}, S_{5}, S_{6}$ are the same as well, let us say $q$. If $p>q$ we make $(p-q)$ times move $k_{456}$, if $q>p$ we make $(q-p)$ times move $k_{123}$ and we are done.

Now let us determine the number of six element sets $M=\left\{c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}\right\}$ of positive integers, such that

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

that is

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

Wlog we may assume $c_{1}<c_{2}<c_{3}$ (and consequently $c_{4}<c_{5}<c_{6}$ ) and because of (2) we have

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

Due to (2), each triple $\left(c_{1}, c_{2}, c_{3}\right)$ uniquely determines $c_{4}, c_{5}$, and $c_{6}$. Thus the number of sought sets is equal to the number of triples $\left(c_{1}, c_{2}, c_{3}\right)$ of positive integers satisfying $c_{1}<c_{2}<c_{3}<10$, which is

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