Problem 1126

## Solution.

From each cube, three sides are visible that meet at one vertex of the large cube, and on them are one, two, or three dots. Then the total number of dots visible on the large cube is $8 \cdot(1+2+3)=$ 48.

Let the number of dots on the sides of the cube be denoted as $\mathrm{s} a_{1}, a_{2}, \ldots a_{6}$ in ascending order. If these are different terms of an arithmetic sequence, then the difference $d>0, d \in \mathbb{N}$.

The sum of the arithmetic sequence of 6 terms is $\frac{6}{2}\left(2 a_{1}+5 d\right)$, so

$\frac{6}{2}\left(2 a_{1}+5 d\right)=48$, or $2 a_{1}+5 d=16$.

From this, it follows that $5 d=2\left(8-a_{1}\right)$, so $d$ must be an even number, and $4 \leqslant a_{1} \leqslant 8$.

By direct verification, it follows that $a_{1}=8, d=0$. Therefore, the number of dots on all six sides of the cube cannot be different terms of an arithmetic sequence.

Note: We reach the same conclusion if we consider the sum of the arithmetic sequence as $\frac{6}{2}\left(a_{1}+a_{6}\right)=$ $48 \Leftrightarrow a_{1}+a_{6}=16$. Then by listing and verifying all possibilities $(4+12,5+11,6+10$, $7+9,8+8$ ), from $a_{6}=a_{1}+5 d$, we find that the difference $d$ is not a natural number.