9.2. Ten numbers are written around a circle, the sum of which is 100. It is known that the sum of any three consecutive numbers is not less than 29. Indicate the smallest number $A$ such that in any set of numbers satisfying the condition, each number does not exceed $A$.
Answer. $A=13$

Solution. Let $X$ be the largest of the listed numbers. The remaining numbers can be divided into 3 "neighbor" triplets. The sum of the numbers in each such triplet is no less than 29, therefore, $X \leq 100 - 3 \cdot 29 = 13$. An example of a set with the maximum number 13: $13,9,10,10,9,10,10,9,10,10$.

Grading criteria. Proven that the desired number is no less than 13: 4 points. Example for 13: 3 points