Problem 4. A row of 11 numbers is written such that the sum of any three consecutive numbers is 18. Additionally, the sum of all the numbers is 64. Find the central number.
Answer: 8.

Solution. Number the numbers from left to right from 1 to 11.

Notice that the sum of the five central numbers (from the fourth to the eighth) is 64 (the sum of all numbers) $-2 \cdot 18$ (the sum of the numbers in the first and last triplets) $=28$.

Then the sixth (central) number is 18 (the sum of the fourth, fifth, and sixth numbers) +18 (the sum of the sixth, seventh, and eighth numbers) -28 (the sum of the central five numbers) $=8$.