11-5. Santa Claus is preparing gifts. He has distributed 115 candies into bags, with each bag containing a different number of candies. In the three smallest gifts, there are 20 candies, and in the three largest - 50. How many bags are the candies distributed into? How many candies are in the smallest gift?
Answer: 10 packages, 5 candies.

Solution. Let's number the gifts from the smallest to the largest, from 1 to n. If the third gift has 7 or fewer candies, then the three smallest gifts have no more than $7+6+5=18$ candies. This contradicts the condition. Therefore, the third gift has at least 8 candies. Similarly, the third from the last gift has no more than 15 candies $(16+17+18=51>50)$.

Remove the three largest and the three smallest gifts. In the remaining gifts, there will be 115 - 20 $50=45$ candies, and each will have between 9 and 14 candies. Three packages are clearly not enough ( $14+13+12=39$ ), and five would be too many ( $9+10+11+12+13=55$ ). Therefore, 45 candies are distributed in 4 packages. This is possible: $47=9+11+12+13$. Note that the fourth package cannot have more than 9 candies: $10+11+12+13=46>45$.

If the fourth package has 9 candies, then the third has no more than 8, the second no more than 7, so the first package has at least $20-8-7=5$ candies. But no more, since $6+7+8=21$.

Criteria. Only the answer - 0 points. Only correct estimates for $a_{3}$ and $a_{n-3}-3$ points. Only a justified answer for the number of packages - 5 points. Full solution - 7 points.