2. Twenty numbers are arranged in a circle. It is known that the sum of any six consecutive numbers is 24. What is the number in the 12th position if the number in the 1st position is 1?
Answer: 7.

Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 20)$. Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:

$$
a_{1}=a_{7}=a_{13}=a_{19}=a_{5}=a_{11}=a_{17}=a_{3}=a_{9}=a_{15}=a_{1}
$$

Now it is clear that all numbers at odd positions are equal to each other. The same is true for numbers at even positions. Therefore, the numbers go like this:

$$
x, y, x, y, \ldots, x, y
$$

From the condition, it follows that

$$
3(x+y)=24, \quad x=1
$$

Hence, $y=7$. Therefore, ones are at odd positions, and sevens are at even positions.

Evaluation. 12 points for a complete solution.