Iranian Mathematical Olympiad

The answer is no. First, for every $1\le i\le 8$ we define $N_i$ to be the set of numbers assigned to the vertices adjacent to the vertex with number $i$. By the problem's assumption, the sum of elements of $N_i$ is divisible by $i$. Since the sum of every three numbers less than $8$ is at most $7+6+5=18$, the sum of elements of $N_8$ is either $8$ or $16$. By checking different cases, it is easy to see $N_8$ equals to either $\{3,6,7\}$, $\{4,5,7\}$, $\{5,2,1\}$, or $\{4,3,1\}$. We will check these different cases.

If $N_8=\{3,6,7\}$, let $N_6=\{x,y,8\}$. Evidently $N_8\cap N_6=\varnothing$. Now, by assumption, $6\mid x+y+8$, and $11=1+2+8\le x+y+8\le 17=4+5+8$. These imply $x+y=4$ and so $\{x,y\}=\{1,3\}$. This contradicts $N_8\cap N_6=\varnothing$.

The case $N_8=\{4,5,7\}$ is similar to the previous case. We only need to consider $N_7$ instead of $N_6$.

If $N_8=\{1,3,4\}$, denote by $x,y,z,t$ the numbers of other vertices as in the figure below. We have $\{x,y,z,t\}=\{2,5,6,7\}$. It follows from $t\mid x+y+z$ that $t$ is a divisor of $t+x+y+z=20$ so $t=5$. On the other hand, $y\mid 1+3+t$. Therefore $y$ must be either $1$, $3$ which is impossible.



If $N_8=\{5,2,1\}$, denote the numbers of other vertices by $x,y,z,t$ as in the figure below. In this case $\{x,y,z,t\}=\{3,4,6,7\}$. By a similar argument to the previous case, we have $t=4$. On the other hand, $x\mid 1+5+t$, so $x$ is a divisor of $10$, again impossible.



Therefore, in each case, the numbers cannot be assigned to the vertices of the cube and the solution is complete. ■