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 such set of numbers, each number does not exceed $A$.
Each number from the given set of numbers is complemented to 100 by three "neighbors" of threes.

## Solution

Let $X$ be the largest of the numbers. The remaining numbers will be divided into three "neighbor" triples. The sum of the numbers in each such triple is at least 29, therefore,

$X \leq 100-3 \cdot 29=13$.

An example of a set with the maximum number $13: 13,9,10,10,9,10,10,9,10,10$.

## Answer

$A=13$.