# 8. Solution.

a) Suppose the set contains 7 numbers. Then four of them are not less than the median, which is the number 10. Another three numbers are not less than one. Then the sum of all numbers in the set is not less than

$$
3+4 \cdot 10=43
$$

Since the arithmetic mean is 6, the sum of the numbers in the set is $6 \cdot 7=42$. Contradiction. The set cannot consist of 7 numbers.

b) Let the number of numbers in the set be $2 n+1$ (where $n$ is a natural number). In the set, there will be exactly $n+1$ numbers, each of which is not less than the median, which is the number 10. The remaining $n$ numbers are not less than the number 1. Then the sum of all numbers in the set is not less than

$$
n+10(n+1)=11 n+10
$$

On the other hand, this sum is equal to $6(2 n+1)=12 n+6$. From the inequality $12 n+6 \geq 11 n+10$ we get that $n \geq 4$. Therefore, the set contains no fewer than $2 \cdot 4+1=9$ numbers. Let's provide an example to show that this is possible. The set

$$
\text { 1; 1; 1; 1; 10; 10; 10; 10; } 10
$$

consists of 9 numbers and satisfies conditions 1-4.
Answer: a) no; b) 9.

## Grading Criteria

| Both parts solved correctly or only part (b) | 3 points |
| :--- | :---: |
| Correct estimate of the number of numbers in part (b), but no example | 2 points |
| Part (a) solved correctly | 1 point |
| Solution is incorrect or missing (including only the answer) | 0 points |