Problem 11.1. Twelve different natural numbers are written in a circle, one of which is 1. Any two adjacent numbers differ by either 10 or 7. What is the greatest value that the largest written number can take?
Answer: 58.

Solution. For convenience, let's number all the numbers in a circle clockwise, starting from the number 1: $a_{1}=1, a_{2}, \ldots, a_{12}$. Notice that for all $1 \leqslant i \leqslant 6$ we have

$$
\begin{gathered}
a_{i+1}-a_{1}=\left(a_{i+1}-a_{i}\right)+\left(a_{i}-a_{i-1}\right)+\ldots+\left(a_{2}-a_{1}\right) \leqslant 10 i \\
a_{13-i}-a_{1}=\left(a_{13-i}-a_{14-i}\right)+\ldots+\left(a_{12}-a_{1}\right) \leqslant 10 i
\end{gathered}
$$

therefore, $a_{7}$ does not exceed $1+6 \cdot 10=61$, and all other numbers do not exceed $1+5 \cdot 10=51$.

If $a_{7}=61$, then all differences $a_{2}-a_{1}, a_{3}-a_{2}, \ldots, a_{7}-a_{6}, a_{7}-a_{8}, a_{8}-a_{9}, \ldots, a_{12}-a_{1}$ are equal to 10, but then $a_{2}=a_{12}=11$, which is a contradiction. Therefore, $a_{7}<61$, which implies that $a_{7} \leqslant 1+5 \cdot 10+7=58$. Thus, the largest written number does not exceed 58.

It remains to check that the number 58 can indeed be present on the board. Let the numbers $1,11,21,31,41,51,58,48,38,28,18,8$ be written in a circle clockwise.

![](https://cdn.mathpix.com/cropped/2024_05_06_20f31b6c2b9751326bf1g-2.jpg?height=77&width=275&top_left_y=991&top_left_x=595)

It is easy to see that all the conditions of the problem are satisfied.